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CHAPTER 13

MORE SPECIAL FUNCTIONS

In this chapter we shall study four sets of orthogonal polynomials, Hermite, Laguerre, andChebyshev1 of first and second kinds. Although these four sets are of less importance inmathematical physics than are the Bessel and Legendre functions of Chapters 11 and 12,they are used and therefore deserve attention. For example, Hermite polynomials occur insolutions of the simple harmonic oscillator of quantum mechanics and Laguerre polynomi-als in wave functions of the hydrogen atom. Because the general mathematical techniquesduplicate those of the preceding two chapters, the development of these functions is onlyoutlined. Detailed proofs, along the lines of Chapters 11 and 12, are left to the reader. Weexpress these polynomials and other functions in terms of hypergeometric and confluenthypergeometric functions. To conclude the chapter, we give an introduction to Mathieufunctions, which arise as solutions of ODEs and PDEs with elliptical boundary conditions.

13.1 HERMITE FUNCTIONS

Generating Functions — Hermite Polynomials

The Hermite polynomials (Fig. 13.1),Hn(x), may be defined by the generating function2

g(x, t)= e−t2+2tx =

∞∑

n=0

Hn(x)tn

n! . (13.1)

1This is the spelling choice of AMS-55 (for the complete reference see footnote 4 in Chapter 5). However, a variety of names,such as Tschebyscheff, is encountered.2A derivation of this Hermite-generating function is outlined in Exercise 13.1.1.

817

818 Chapter 13 More Special Functions

FIGURE 13.1 Hermitepolynomials.

Recurrence Relations

Note the absence of a superscript, which distinguishes Hermite polynomials from the unre-lated Hankel functions. From the generating function we find that the Hermite polynomialssatisfy the recurrence relations

Hn+1(x)= 2xHn(x)− 2nHn−1(x) (13.2)

and

H ′n(x)= 2nHn−1(x). (13.3)

Equation (13.2) is obtained by differentiating the generating function with respect tot :

∂g

∂t= (−2t + 2x)e−t

2+2tx =∞∑

n=0

Hn+1(x)tn

n!

= −2∞∑

n=0

Hn(x)tn+1

n! + 2x∞∑

n=0

Hn(x)tn

n! ,

which can be rewritten as∞∑

n=0

tn

n![Hn+1(x)− 2xHn(x)+ 2nHn−1(x)

]= 0.

Because each coefficient of this power series vanishes, Eq. (13.2) is established. Similarly,differentiation with respect tox leads to

∂g

∂x= 2te−t

2+2tx =∞∑

n=0

H ′n(x)

tn

n! = 2∞∑

n=0

Hn(x)tn+1

n! ,

which yields Eq. (13.3) upon shifting the summation indexn in the last sumn+ 1→ n.

13.1 Hermite Functions 819

Table 13.1 Hermite Polynomials

H0(x)= 1H1(x)= 2xH2(x)= 4x2− 2H3(x)= 8x3− 12xH4(x)= 16x4− 48x2+ 12H5(x)= 32x5− 160x3+ 120xH6(x)= 64x6− 480x4+ 720x2− 120

The Maclaurin expansion of the generating function

e−t2+2tx =

∞∑

n=0

(2tx − t2)n

n! = 1+(2tx − t2

)+ · · · (13.4)

givesH0(x)= 1 andH1(x)= 2x, and then the recursion Eq. (13.2) permits the construc-tion of anyHn(x) desired (integraln). For convenient reference the first several Hermitepolynomials are listed in Table 13.1.

Special values of the Hermite polynomials follow from the generating function forx = 0:

e−t2 =

∞∑

n=0

(−t2)nn! =

∞∑

n=0

Hn(0)tn

n! ,

that is,

H2n(0)= (−1)n(2n)!n! , H2n+1(0)= 0, n= 0,1, . . . . (13.5)

We also obtain from the generating function the important parity relation

Hn(x)= (−1)nHn(−x) (13.6)

by noting that Eq. (13.1) yields

g(−x,−t)=∞∑

n=0

Hn(−x)(−t)nn! = g(x, t)=

∞∑

n=0

Hn(x)tn

n! .

Alternate Representations

The Rodrigues representation ofHn(x) is

Hn(x)= (−1)nex2 dn

dxne−x

2. (13.7)

Let us show this using mathematical induction as follows.


Example 13.1.1 RODRIGUES REPRESENTATION

We rewrite the generating function asg(x, t)= ex2e−(t−x)

2and note that

∂

∂te−(t−x)

2 =− ∂

∂xe−(t−x)

2.

This yields

∂g

∂t

∣∣∣∣t=0= (2x − 2t)g

∣∣∣t=0= 2x =H1(x)=−ex

2 d

dxe−x

2,

which is the initialn= 1 case. Assuming the casen of Eq. (13.7) as valid, we now use theoperator identityd

dxex

2 = 2xex2 + ex

2 ddx

in

(−1)n+1ex2 dn+1

dxn+1e−x

2 = (−1)n+1[d

dxex

2 − 2xex2]dn

dxne−x

2

= − d

dxHn(x)+ 2xHn(x)=Hn+1(x)

to establish then+ 1 case, with the last equality following from Eqs.(13.2) and (13.3).More directly, differentiation of the generating functionn times with respect tot and

then settingt equal to zero yields

Hn(x)=∂n

∂tn

(e−t

2+2tx)∣∣∣t=0= (−1)nex

2 ∂n

∂xne−(t−x)

2∣∣∣t=0= (−1)nex

2 dn

dxne−x

2.

�

A second representation may be obtained by using the calculus of residues (Section 7.1).If we multiply Eq. (13.1) byt−m−1 and integrate around the origin in the complext-plane,only the term withHm(x) will survive:

Hm(x)=m!2πi

∮t−m−1e−t

2+2tx dt. (13.8)

Also, from the Maclaurin expansion, Eq. (13.4), we can derive our Hermite polynomialHn(x) in series form: Using the binomial expansion of(2x− t)ν and the indexN = s+ ν,

e−t2+2tx =

∞∑

ν=0

tν

ν! (2x − t)ν =∞∑

ν=0

tν

ν!ν∑

s=0

(ν

s

)(2x)ν−s(−t)s

=∞∑

N=0

tN

N !

[N/2]∑

s=0

(2x)N−2s(−1)sN !

(N − s)!

(N − s

s

),

where[N/2] is the largest integer less than or equal toN/2. Writing the binomial coeffi-cient in terms of factorials and using Eq. (13.1) we obtain

HN (x)=[N/2]∑

s=0

(2x)N−2s(−1)sN !

s!(N − 2s)! .


More explicitly, replacingN→ n, we have

Hn(x) = (2x)n − 2n!(n− 2)!2! (2x)

n−2+ 4n!(n− 4)!4! (2x)

n−41 · 3 · · ·

=[n/2]∑

s=0

(−2)s(2x)n−2s(

n

2s

)1 · 3 · 5 · · · (2s − 1)

=[n/2]∑

s=0

(−1)s(2x)n−2s n!(n− 2s)!s! . (13.9)

This series terminates for integraln and yields our Hermite polynomial.

Orthogonality

If we substitute the recursion Eq. (13.3) into Eq. (13.2) we can eliminate the indexn− 1,obtaining

Hn+1(x)= 2xHn(x)−H ′n(x),

which was used already in Example 13.1.1. If we differentiate this recursion relation andsubstitute Eq. (13.3) for the indexn+ 1 we find

H ′n+1(x)= 2(n+ 1)Hn(x)= 2Hn(x)+ 2xH ′

n(x)−H ′′n (x),

which can be rearranged to the second-order ODE for Hermite polynomials. Thus, therecurrence relations (Eqs. (13.2) and (13.3)) lead to the second-order ODE

H ′′n (x)− 2xH ′

n(x)+ 2nHn(x)= 0, (13.10)

which is clearlynot self-adjoint.To put the ODE in self-adjoint form, following Section 10.1, we multiply by exp(−x2),

Exercise 10.1.2. This leads to the orthogonality integral∫ ∞

−∞Hm(x)Hn(x)e

−x2dx = 0, m = n, (13.11)

with the weighting function exp(−x2), a consequence of putting the ODE into self-adjointform. The interval(−∞,∞) is chosen to obtain the Hermitian operator boundary condi-tions, Section 10.1. It is sometimes convenient to absorb the weighting function into theHermite polynomials. We may define

ϕn(x)= e−x2/2Hn(x), (13.12)

with ϕn(x) no longer a polynomial.Substitution into Eq. (13.10) yields the differential equation forϕn(x),

ϕ′′n(x)+(2n+ 1− x2)ϕn(x)= 0. (13.13)

This is the differential equation for a quantum mechanical, simple harmonic oscillator,which is perhaps the most important physics application of the Hermite polynomials.


Equation (13.13) is self-adjoint, and the solutionsϕn(x) are orthogonal for the interval(−∞< x <∞) with a unit weighting function.

The problem of normalizing these functions remains. Proceeding as in Section 12.3, wemultiply Eq. (13.1) by itself and then bye−x

2. This yields

e−x2e−s

2+2sxe−t2+2tx =

∞∑

m,n=0

e−x2Hm(x)Hn(x)

smtn

m!n! .

When we integrate this relation overx from −∞ to +∞, the cross terms of the doublesum drop out because of the orthogonality property:3

∞∑

n=0

(st)n

n!n!

∫ ∞

−∞e−x

2[Hn(x)

]2dx =

∫ ∞

−∞e−x

2−s2+2sx−t2+2tx dx

=∫ ∞

−∞e−(x−s−t)

2e2st dx

= π1/2e2st = π1/2∞∑

n=0

2n(st)n

n! , (13.14)

using Eqs. (8.6) and (8.8). By equating coefficients of like powers ofst , we obtain∫ ∞

−∞e−x

2[Hn(x)

]2dx = 2nπ1/2n!. (13.15)

Quantum Mechanical Simple Harmonic Oscillator

The following development of Hermite polynomials via simple harmonic oscillator wavefunctionsφn(x) is analogous to the use of the raising and lowering operators for angu-lar momentum operators presented in Section 4.3. This means that we derive the eigen-valuesn + 1/2 and eigenfunctions (theHn(x)) without assuming the development that

led to Eq. (13.13). The key aspect of the eigenvalue Eq. (13.13),( d2

dx2 − x2)ϕn(x) =−(2n+ 1)ϕn(x), is that the Hamiltonian

−2H≡ d2

dx2− x2=

(d

dx− x

)(d

dx+ x

)+[x,

d

dx

](13.16)

almost factorizes. Using naivelya2−b2= (a−b)(a+b), the basic commutator[px, x] =h/i of quantum mechanics (with momentumpx = (h/i)d/dx) enters as a correction inEq. (13.16). (Becausepx is Hermitian,d/dx is anti-Hermitian,(d/dx)†=−d/dx.) Thiscommutator can be evaluated as follows. Imagine the differential operatord/dx acts on awave functionϕ(x) to the right, as in Eq. (13.13), so

d

dx(xϕ)= x

d

dxϕ + ϕ, (13.17)

3The cross terms(m = n) may be left in, if desired. Then, when the coefficients ofsα tβ are equated, the orthogonality will beapparent.


by the product rule. Dropping the wave functionϕ from Eq. (13.17), we rewrite Eq. (13.17)as

d

dxx − x

d

dx≡[d

dx, x

]= 1, (13.18)

a constant, and then verify Eq. (13.16) directly by expanding the product of operators.The product form of Eq. (13.16), up to the constant commutator, suggests introducing thenon-Hermitian operators

a†≡ 1√2

(x − d

dx

), a ≡ 1√

2

(x + d

dx

), (13.19)

with (a)†= a†, which are adjoints of each other. They obey the commutation relations

[a, a†]=

[d

dx, x

]= 1, [a, a] = 0=

[a†, a†], (13.20)

which are characteristic of these operators and straightforward to derive from Eq. (13.18)and

[d

dx,d

dx

]= 0= [x, x] and

[x,

d

dx

]=−

[d

dx, x

].

Returning to Eq. (13.16) and using Eq. (13.19) we rewrite the Hamiltonian as

H= a†a + 1

2= a†a + 1

2

(a†a + aa†)= 1

2

(a†a + aa†) (13.21)

and introduce the Hermitiannumber operator N = a†a so thatH=N + 1/2. Let |n〉 bean eigenfunction ofH,

H|n〉 = λn|n〉,whose eigenvalueλn is unknown at this point. Now we prove the key property thatN hasnonnegative integer eigenvalues

N |n〉 =(λn −

1

2

)|n〉 = n|n〉, n= 0,1,2 . . . , (13.22)

that is,λn = n+ 1/2. Sincea|n〉 is complex conjugate to〈n|a†, the normalization integral〈n|a†a|n〉 ≥ 0 and is finite. From

(a|n〉

)†a|n〉 = 〈n|a†a|n〉 =

(λn −

1

2

)≥ 0 (13.23)

we see thatN has nonnegative eigenvalues.We now show that ifa|n〉 is nonzero it is an eigenfunction with eigenvalueλn−1 =

λn − 1. After normalizinga|n〉, this state is designated|n − 1〉. This is proved by thecommutation relations

[N, a†]= a†, [N, a] = −a, (13.24)


which follow from Eq. (13.20). These commutation relations characterizeN as the numberoperator. To see this, we determine the eigenvalue ofN for the statesa†|n〉 anda|n〉. Usingaa†=N + [a, a†] =N + 1, we find that

N(a†|n〉

)= a†(aa†)|n〉 = a†([a, a†]+N

)|n〉

= a†(N + 1)|n〉 =(λn +

1

2

)a†|n〉 = (n+ 1)a†|n〉, (13.25)

N(a|n〉

)=(aa†− 1

)a|n〉 = a(N − 1)|n〉 = (n− 1)a|n〉.

In other words,N acting ona†|n〉 shows thata† has raised the eigenvaluen correspondingto |n〉 by one unit, whence its nameraising, orcreation, operator. Applying a† repeatedly,we can reach all higher states. There is no upper limit to the sequence of eigenvalues.Similarly, a lowers the eigenvaluen by one unit; hence it is alowering (or annihilation )operator because. Therefore,

a†|n〉 ∼ |n+ 1〉, a|n〉 ∼ |n− 1〉. (13.26)

Applying a repeatedly, we can reach the lowest, or ground, state|0〉with eigenvalueλ0. Wecannot step lower becauseλ0≥ 1/2. Thereforea|0〉 ≡ 0, suggesting we constructψ0= |0〉from the (factored)first-order ODE

√2aψ0=

(d

dx+ x

)ψ0= 0. (13.27)

Integrating

ψ ′0ψ0=−x, (13.28)

we obtain

lnψ0=−1

2x2+ ln c0, (13.29)

wherec0 is an integration constant. The solution,

ψ0(x)= c0e−x2/2, (13.30)

is a Gaussian that can be normalized, withc0= π−1/4 using the error integral, Eqs. (8.6)and (8.8). Substitutingψ0 into Eq. (13.13) we find

H|0〉 =(a†a + 1

2

)|0〉 = 1

2|0〉, (13.31)

so its energy eigenvalue isλ0 = 1/2 and its number eigenvalue isn = 0, confirming thenotation|0〉. Applying a† repeatedly toψ0= |0〉, all other eigenvalues are confirmed to beλn = n+1/2, proving Eq. (13.13). The normalizations needed for Eq. (13.26) follow fromEqs. (13.25) and (13.23) and

〈n|aa†|n〉 = 〈n|a†a + 1|n〉 = n+ 1, (13.32)


showing

√n+ 1|n+ 1〉 = a†|n〉, √

n|n− 1〉 = a|n〉. (13.33)

Thus, the excited-state wave functions,ψ1,ψ2, and so on, are generated by the raisingoperator

|1〉 = a†|0〉 = 1√2

(x − d

dx

)ψ0(x)=

x√

2

π1/4e−x

2/2, (13.34)

yielding (and leading to upcoming Eq. (13.38))

ψn(x)=NnHn(x)e−x2/2, Nn ≡ π−1/4(2nn!

)−1/2, (13.35)

whereHn are the Hermite polynomials (Fig. 13.2).As shown, the Hermite polynomials are used in analyzing the quantum mechanical sim-

ple harmonic oscillator. For a potential energyV = 12Kz2= 1

2mω2z2 (forceF=−∇V =−Kzz), the Schrödinger wave equation is

− h2

2m∇2�(z)+ 1

2Kz2�(z)=E�(z). (13.36)

Our oscillating particle has massm and total energyE. By use of the abbreviations

x = αz with α4= mK

h2= m2ω2

h2,

λ= 2E

h

(m

K

)1/2

= 2E

hω,

(13.37)

in whichω is the angular frequency of the corresponding classical oscillator, Eq. (13.36)becomes (with�(z)=�(x/α)=ψ(x))

d2ψ(x)

dx2+(λ− x2)ψ(x)= 0. (13.38)

This is Eq. (13.13) withλ= 2n+ 1. Hence (Fig. 13.2),

ψn(x)= 2−n/2π−1/4(n!)−1/2e−x2/2Hn(x), (normalized). (13.39)

Alternatively, the requirement thatn be an integer is dictated by the boundary conditionsof the quantum mechanical system,

limz→±∞

�(z)= 0.

Specifically, ifn→ ν, not an integer, a power-series solution of Eq. (13.13) (Exercise 9.5.6)shows thatHν(x) will behave asxνex

2for largex. The functionsψν(x) and�ν(z) will

therefore blow up at infinity, and it will be impossible to normalize the wave function�(z).With this requirement, the energyE becomes

E =(n+ 1

2

)hω. (13.40)


FIGURE 13.2 Quantum mechanicaloscillator wave functions: The

heavy bar on thex-axis indicates theallowed range of the classical

oscillator with the same total energy.

As n ranges over integral values(n≥ 0), we see that the energy is quantized and that thereis a minimum or zero point energy

Emin=1

2hω. (13.41)

This zero point energy is an aspect of the uncertainty principle, a genuine quantum phe-nomenon.


In quantum mechanical problems, particularly in molecular spectroscopy, a number ofintegrals of the form

∫ ∞

−∞xre−x

2Hn(x)Hm(x) dx

are needed. Examples forr = 1 andr = 2 (with n=m) are included in the exercises at theend of this section. A large number of other examples are contained in Wilson, Decius, andCross.4

In the dynamics and spectroscopy of molecules in the Born–Oppenheimer approxima-tion, the motion of a molecule is separated into electronic, vibrational and rotational mo-tion. Each vibrating atom contributes to a matrix element two Hermite polynomials, itsinitial state and another one to its final state. Thus, integrals of products of Hermite poly-nomials are needed.

Example 13.1.2 THREEFOLD HERMITE FORMULA

We want to calculate the following integral involvingm= 3 Hermite polynomials:

I3≡∫ ∞

−∞e−x

2HN1(x)HN2(x)HN3(x) dx, (13.42)

whereNi ≥ 0 are integers. The formula (due to E. C. Titchmarsh,J. Lond. Math. Soc.23:15 (1948), see Gradshteyn and Ryzhik, p. 838, in the Additional Readings) generalizes them = 2 case needed for the orthogonality and normalization of Hermite polynomials. Toderive it, we start with the product of three generating functions of Hermite polynomials,multiply by e−x

2, and integrate overx in order to generateI3:

Z3 ≡∫ ∞

−∞e−x

23∏

j=1

e2xtj−t2j dx =

∫ ∞

−∞e−(∑3

j=1 tj−x)2+2(t1t2+t1t3+t2t3) dx

=√πe2(t1t2+t1t3+t2t3). (13.43)

The last equality follows from substitutingy = x −∑j tj and using the error integral∫∞−∞ e−y

2dy =√π , Eqs. (8.6) and (8.8). Expanding the generating functions in terms of

Hermite polynomials we obtain

Z3 =∞∑

N1,N2,N3=0

tN11 t

N22 t

N33

N1!N2!N3!

∫ ∞

−∞e−x

2HN1(x)HN2(x)HN3(x) dx

=√π

∞∑

N=0

2N

N ! (t1t2+ t1t3+ t2t3)N

=√π

∞∑

N=0

2N

N !∑

0≤ni≤N,∑

i ni=N

N !n1!n2!n3!

(t1t2)n1(t1t3)

n2(t2t3)n3,

4E. B. Wilson, Jr., J. C. Decius, and P. C. Cross,Molecular Vibrations, New York: McGraw-Hill (1955), reprinted Dover (1980).


using the polynomial expansion

( m∑

j=1

aj

)N

=∑

0≤ni≤m

N !n1! · · ·nm!

an11 · · ·anmm .

The powers of the foregoingtj tk become

(t1t2)n1(t1t3)

n2(t2t3)n3 = t

N11 t

N22 t

N33 ;

N1= n1+ n2, N2= n1+ n3, N3= n2+ n3.

That is, from

2N = 2(n1+ n2+ n3)=N1+N2+N3

there follows

2N = 2n1+ 2N3= 2n2+ 2N2= 2n3+ 2N1,

so we obtain

n1=N −N3, n2=N −N2, n3=N −N1.

The ni are all fixed (making this case special and easy) because theNi are fixed, and

2N =3∑

i=1Ni , with N ≥ 0 an integer by parity. Hence, upon comparing the foregoing like

t1t2t3 powers,

I3=√π2NN1!N2!N3!

(N −N1)!(N −N2)!(N −N3)!, (13.44)

which is the desired formula. If weorder N1 ≥ N2 ≥ N3 ≥ 0, thenn1 ≥ n2 ≥ n3 ≥ 0follows, being equivalent toN −N3≥N −N2≥N −N1≥ 0, which occur in the denom-inators of the factorials ofI3. �

Example 13.1.3 DIRECT EXPANSION OF PRODUCTS OF HERMITE POLYNOMIALS

In an alternative approach, we now start again from the generating function identity

∞∑

N1,N2=0

HN1(x)HN2(x)tN11

N1!tN22

N2!= e2x(t1+t2)−t21−t22 = e2x(t1+t2)−(t1+t2)2 · e2t1t2

=∞∑

N=0

HN (x)(t1+ t2)

N

N !

∞∑

ν=0

(2t1t2)ν

ν! .


Using the binomial expansion and then comparing like powers oft1t2 we extract an identitydue to E. Feldheim (J. Lond. Math. Soc.13: 22 (1938)):

HN1(x)HN2(x) =min(N1,N2)∑

ν=0

HN1+N2−2νN1!N2!2ν

ν!(N1+N2− 2ν)!

(N1+N2− 2ν

N1− ν

)

=∑

0≤ν≤min(N1,N2)

HN1+N2−2ν2νν!(N1ν

)(N2ν

). (13.45)

For ν = 0 the coefficient ofHN1+N2 is obviously unity. Special cases, such as

H 21 =H2+2, H1H2=H3+4H1, H 2

2 =H4+8H2+8, H1H3=H4+6H2,

can be derived from Table 13.1 and agree with the general twofold product formula.This compact formula can be generalized to products ofm Hermite polynomials, and

this in turn yields a new closed form result for the integralIm.Let us begin with a new result forI4 containing a product of four Hermite polynomials.

Inserting the Feldheim identity forHN1HN2 andHN3HN4 and using orthogonality∫ ∞

−∞e−x

2HN1HN2 dx =

√π2N1N1!δN1N2

for the remaining product of two Hermite polynomials yields

I4 =∫ ∞

−∞e−x

2HN1HN2HN3HN4 dx

=∑

0≤µ≤min(N1,N2);0≤ν≤min(N3,N4)

2µµ!

·(N1µ

)(N2µ

)2νν!

(N3ν

)(N4ν

)∫ ∞

−∞e−x

2HN1+N2−2µHN3+N4−2ν dx

=N4∑

ν=0

√π2M(N3+N4− 2ν)!N1!N2!N3!N4!

(M −N3−N4− ν)!(M −N1+ ν)!(M −N2+ ν)!(N3− ν)!(N4− ν)!ν! .

(13.46)

Here we use the notationM = (N1+N2+N3+N4)/2 and write the binomial coefficientsexplicitly, so

12(N1+N2−N3−N4) =M −N3−N4,

12(N1−N2+N3+N4) =M −N2,

12(N2−N1+N3+N4) =M −N1.

From orthogonality we haveµ = (N1 + N2 − N3 − N4)/2 + ν. The upper limitof ν is min(N3,N4,M − N1,M − N2) = min(N4,M − N1) and the lower limit ismax(0,N3+N4−M)= 0, if we orderN1≥N2≥N3≥N4.


Now we return to the product expansion ofm Hermite polynomials and the correspond-ing new result from it forIm. We prove a generalized Feldheim identity,

HN1(x) · · ·HNm(x)=∑

ν1,...,νm−1

HM(x)aν1,...,νm−1, (13.47)

where

M =m−1∑

i=1

(Ni − 2νi)+Nm,

by mathematical induction. Multiplying this byHNm+1 and using the Feldheim identity,we end up with the same formula form+ 1 Hermite polynomials, including the recursionrelation

aν1,...,νm = aν1,...,νm−12νmνm!(Nm+1νm

)(∑m−1i=1 (Ni − 2νi)+Nm+1

νm

).

Its solution is

aν1,...,νm−1 =m−1∏

i=1

(Ni+1νi

)(∑i−1j=1(Nj − 2νj )+Ni

νi

)2νiνi !. (13.48)

The limits of the summation indices are

0≤ ν1≤min(N1,N2), 0≤ ν2≤min(N3,N1+N2− 2ν1), . . . ,

0≤ νm−1≤min

(Nm,

m−2∑

i=1

(Ni − 2νi)+Nm−1

). (13.49)

We now apply this generalized Feldheim identity, with indices ordered asN1 ≥ N2 ≥· · · ≥ Nm, to Im, groupingHN2 · · ·HNm together and using orthogonality for the re-maining product of two Hermite polynomialsHN1H

∑m−1i=2 (Ni−2νi )+Nm

. This yieldsN1 =∑m−1

i=2 (Ni − 2νi)+Nm, fixing νm−1, and

Im =√π2N1N1!

∑

ν2,...,νm−1

m−1∏

i=2

(Ni+1νi

)(∑i−1j=2(Nj − 2νj )+Ni

νi

)νi !2νi , (13.50)

where the limits on the summation indices are

0≤ ν2≤min(N3,N2), . . . , 0≤ νm−1≤min

(Nm,

m−2∑

i=2

(Ni−2νi)+Nm−1

). (13.51)

�


Example 13.1.4 APPLICATIONS OF THE PRODUCT FORMULAS

To check the expressionIm for m= 3, we note that the sum∑i−1

j=2 with i− 1=m− 2= 1in the second binomial coefficient inIm is empty (that is, zero), so onlyNi =Nm−1=N2

remains. Also, withνm−2 = ν1 the sum over theνi is that overν2, which is fixed by theconstraint on the summation indexν2: N1 = N2 − 2ν2 + N3. Henceν2 = (N2 + N3 −N1)/2= N −N1, with N = (N1+N2+N3)/2. That is, only the product remains inIm.The general formula forIm therefore gives

I3=√π2N1N1!

(N3

ν2

)(N2

ν2

)ν2!2ν2 =

√π2NN1!N2!N3!

(N −N1)!(N −N2)!(N −N3)!,

which agrees with our earlier result of Example 13.1.2. The last expression is based onthe following observations. The power of 2 has the exponentN1+ ν2=N . The factorialsfrom the binomial coefficients areN3 − ν2 = (N1 + N3 − N2)/2= N − N2,N2 − ν2 =(N1+N2−N3)/2=N −N3.

Next let us considerm = 4, where we do not order the Hermite indicesNi as yet. Thereason is that the generalIm expression was derived with a different grouping of the Her-mite polynomials than the separate calculation ofI4 with which we compare. That is whywe’ll have to permute the indices to get the earlier result forI4. That is a general conclu-sion: Different groupings of the Hermite polynomials just give different permutations ofthe Hermite indices in the general result.

We have two summations overν2 andνm−1= ν3, which is fixed by the constraintN1=N2− 2ν2+N3− 2ν3+N4. Hence

ν3= 12(N2+N3+N4−N1)− ν2=M −N1− ν2

with M = 12(N1+N2+N3+N4). The exponent of 2 isN1+ ν2+ ν3=M . Therefore for

m= 4 theIm formula gives

I4 =√π2N1N1!

∑

ν2≥0

(N3

ν2

)(N4

ν3

)(N2

ν2

)(N2− 2ν2+N3

ν3

)ν2!2ν2ν3!2ν3

=∑

ν2≥0

√π2MN1!N2!N3!N4!(N2− 2ν2+N3)!

ν2!ν3!(N2− ν2)!(N3− ν2)!(N4− ν3)!(N2+N3− 2ν2− ν3)!

=∑

ν2≥0

√π2MN1!N2!N3!N4!(N2+N3− 2ν2)!

(N2− ν2)!(N3− ν2)!(N4− ν3)!ν2!ν3!(N2+N3− 2ν2− ν3)!

=∑

ν2≥0

√π2MN1!N2!N3!N4!(N2+N3− 2ν2)!

ν2!(M −N1− ν2)!(N3− ν2)!(N2− ν2)!(M −N2−N3+ ν2)!(M −N4− ν2)!.

In the last expression we have substitutedν3 and used

N4− ν3= (N1−N2−N3+N4)+ ν2=M −N2−N3+ ν2,

N2+N3− 2ν2− ν3=N1+N2+N3−N4

2− ν2=M −N4− ν2.


The upper limit isν2 ≤ min(N2,N3,M − N1,M − N4), and the lower limit isν2 ≥max(0,N2 + N3 −M). If we make the permutationN2 ↔ N4, ν2 → ν, then our previ-ousI4 result is obtained with upper limitν ≤min(N4,M −N1)= 1

2(N2+N3+N4−N1)

and lower limit ν ≥ max(0,N3 + N4 − M) = 0 becauseN3 + N4 − N1 − N2 ≤ 0 forN1≥N2≥N3≥N4≥ 0. �

The Hermite polynomial product formula also applies to products of simple harmonicoscillator wave functions,

∫∞−∞ e−mx2/2HN1(x) · · ·HNm(x) dx, with a different exponential

weight function. To evaluate such integrals we use the generalized Feldheim identity forHN2 · · ·HNm in conjunction with the integral (see Gradshteyn and Ryzhik, p. 845, in theAdditional Readings),

∫ ∞

−∞e−a

2x2Hm(x)Hn(x) dx =

1

2

(2

a

)m+n+1(1− a2)(m+n)/2

Ŵ

(m+ n+ 1

2

)

· 2F1

(−m,−n; 1−m− n

2; a2

2(a2− 1)

),

instead of the standard orthogonality integral for the remaining product of two Hermitepolynomials. Here the hypergeometric function is the finite sum

2F1

(−m,−n; 1−m− n

2; a2

2(a2− 1)

)=

min(m,n)−1∑

ν=0

(−m)ν(−n)νν!(1−m−n

2 )ν

(a2

2(a2− 1)

)ν

with (−m)ν = (−m)(1−m) · · · (ν−1−m) and(−m)0≡ 1. This yields a result similar toIm but somewhat more complicated.

The oscillator potential has also been employed extensively in calculations of nuclearstructure (nuclear shell model) quark models of hadrons and the nuclear force.

There is a second independent solution of Eq. (13.13). This Hermite function of thesecond kind is an infinite series (Sections 9.5 and 9.6) and is of no physical interest, atleast not yet.

Exercises

13.1.1 Assume the Hermite polynomials are known to be solutions of the differential equa-tion (13.13). From this the recurrence relation, Eq. (13.3), and the values ofHn(0) arealso known.

(a) Assume the existence of a generating function

g(x, t)=∞∑

n=0

Hn(x)tn

n! .

(b) Differentiateg(x, t) with respect tox and using the recurrence relation develop afirst-order PDE forg(x, t).

(c) Integrate with respect tox, holdingt fixed.(d) Evaluateg(0, t) using Eq. (13.5). Finally, show that

g(x, t)= exp(−t2+ 2tx

).


13.1.2 In developing the properties of the Hermite polynomials, start at a number of differentpoints, such as:

1. Hermite’s ODE, Eq. (13.13),2. Rodrigues’ formula, Eq. (13.7),3. Integral representation, Eq. (13.8),4. Generating function, Eq. (13.1),5. Gram–Schmidt construction of a complete set of orthogonal polynomials over

(−∞,∞) with a weighting factor of exp(−x2), Section 10.3.

Outline how you can go from any one of these starting points to all the other points.

13.1.3 Prove that(

2x − d

dx

)n

1=Hn(x).

Hint. Check out the first couple of examples and then use mathematical induction.

13.1.4 Prove that∣∣Hn(x)

∣∣≤∣∣Hn(ix)

∣∣.

13.1.5 Rewrite the series form ofHn(x), Eq. (13.9), as anascendingpower series.

ANS.H2n(x)= (−1)nn∑

s=0

(−1)2s(2x)2s(2n)!

(2s)!(n− s)! ,

H2n+1(x)= (−1)nn∑

s=0

(−1)s(2x)2s+1 (2n+ 1)!(2s + 1)!(n− s)! .

13.1.6 (a) Expandx2r in a series of even-order Hermite polynomials.(b) Expandx2r+1 in a series of odd-order Hermite polynomials.

ANS. (a)x2r = (2r)!22r

r∑

n=0

H2n(x)

(2n)!(r − n)!

(b) x2r+1= (2r + 1)!22r+1

r∑

n=0

H2n+1(x)

(2n+ 1)!(r − n)! , r = 0,1,2, . . . .

Hint. Use a Rodrigues representation and integrate by parts.

13.1.7 Show that

(a)∫ ∞

−∞Hn(x)exp

[−x2

2

]dx =

{2πn!/(n/2)!, n even0, n odd.

(b)∫ ∞

−∞xHn(x)exp

[−x2

2

]dx =

0, n even

2π(n+ 1)!

((n+ 1)/2)! , n odd.


13.1.8 Show that∫ ∞

−∞xme−x

2Hn(x) dx = 0 for m an integer, 0≤m≤ n− 1.

13.1.9 The transition probability between two oscillator statesm andn depends on∫ ∞

−∞xe−x

2Hn(x)Hm(x) dx.

Show that this integral equalsπ1/22n−1n!δm,n−1 + π1/22n(n + 1)!δm,n+1. This resultshows that such transitions can occur only between states of adjacent energy levels,m= n± 1.Hint. Multiply the generating function (Eq. (13.1)) by itself using two different sets ofvariables(x, s) and(x, t). Alternatively, the factorx may be eliminated by the recur-rence relation, Eq. (13.2).


−∞x2e−x

2Hn(x)Hn(x) dx = π1/22nn!

(n+ 1

2

).

This integral occurs in the calculation of the mean-square displacement of our quantumoscillator.Hint. Use the recurrence relation, Eq. (13.2), and the orthogonality integral.

13.1.11 Evaluate∫ ∞

−∞x2e−x

2Hn(x)Hm(x) dx

in terms ofn andm and appropriate Kronecker delta functions.

ANS. 2n−1π1/2(2n+ 1)n!δnm + 2nπ1/2(n+ 2)!δn+2,m + 2n−2π1/2n!δn−2,m.


−∞xre−x

2Hn(x)Hn+p(x) dx =

{0, p > r

2nπ1/2(n+ r)!, p = r,

with n, p, andr nonnegative integers.Hint. Use the recurrence relation, Eq. (13.2),p times.

13.1.13 (a) Using the Cauchy integral formula, develop an integral representation ofHn(x)

based on Eq. (13.1) with the contour enclosing the pointz=−x.

ANS.Hn(x)=n!

2πiex

2∮

e−z2

(z+ x)n+1dz.

(b) Show by direct substitution that this result satisfies the Hermite equation.


13.1.14 With

ψn(x)= e−x2/2 Hn(x)

(2nn!π1/2)1/2,

verify that

anψn(x) =1√2

(x + d

dx

)ψn(x)= n1/2ψn−1(x),

a†nψn(x) =

1√2

(x − d

dx

)ψn(x)= (n+ 1)1/2ψn+1(x).

Note. The usual quantum mechanical operator approach establishes these raising andlowering properties before the form ofψn(x) is known.

13.1.15 (a) Verify the operator identity

x − d

dx=−exp

[x2

2

]d

dxexp

[−x2

2

].

(b) The normalized simple harmonic oscillator wave function is

ψn(x)=(π1/22nn!

)−1/2 exp

[−x2

2

]Hn(x).

Show that this may be written as

ψn(x)=(π1/22nn!

)−1/2(x − d

dx

)n

exp

[−x2

2

].

Note. This corresponds to ann-fold application of the raising operator of Exer-cise 13.1.14.

13.1.16 (a) Show that the simple oscillator Hamiltonian (from Eq. (13.38)) may be written as

H=−1

2

d2

dx2+ 1

2x2= 1

2

(aa†+ a†a

).

Hint. ExpressE in units of hω.(b) Using the creation–annihilation operator formulation of part (a), show that

Hψ(x)=(n+ 1

2

)ψ(x).

This means the energy eigenvalues areE = (n + 12)(hω), in agreement with

Eq. (13.40).

13.1.17 Write a program that will generate the coefficientsas , in the polynomial form of theHermite polynomialHn(x)=

∑ns=0asx

s .

13.1.18 A functionf (x) is expanded in a Hermite series:

f (x)=∞∑

n=0

anHn(x).


From the orthogonality and normalization of the Hermite polynomials the coefficientan is given by

an =1

2nπ1/2n!

∫ ∞

−∞f (x)Hn(x)e

−x2dx.

Forf (x)= x8 determine the Hermite coefficientsan by the Gauss–Hermite quadrature.Check your coefficients against AMS-55, Table 22.12 (for the reference see footnote 4in Chapter 5 or the General References at book’s end).

13.1.19 (a) In analogy with Exercise 12.2.13, set up the matrix of even Hermite polynomialcoefficients that will transform an even Hermite series into an even power series:

B=

1 −2 12 · · ·0 4 −48 · · ·0 0 16 · · ·...

...... · · ·

.

ExtendB to handle an even polynomial series throughH8(x).(b) Invert your matrix to obtain matrixA, which will transform an even power series

(throughx8) into a series of even Hermite polynomials. Check the elements ofAagainst those listed in AMS-55 (Table 22.12, in the General References at book’send).

(c) Finally, using matrix multiplication, determine the Hermite series equivalent tof (x)= x8.

13.1.20 Write a subroutine that will transform a finite power series,∑N

n=0anxn, into a Hermite

series,∑N

n=0bnHn(x). Use the recurrence relation, Eq. (13.2).Note. Both Exercises 13.1.19 and 13.1.20 are faster and more accurate than the Gaussianquadrature, Exercise 13.1.18, iff (x) is available as a power series.

13.1.21 Write a subroutine for evaluating Hermite polynomial matrix elements of the form

Mpqr =∫ ∞

−∞Hp(x)Hq(x)x

re−x2dx,

using the 10-point Gauss–Hermite quadrature (forp + q + r ≤ 19). Include a paritycheck and set equal to zero the integrals with odd-parity integrand. Also, check to seeif r is in the range|p − q| ≤ r . OtherwiseMpqr = 0. Check your results against thespecific cases listed in Exercises 13.1.9, 13.1.10, 13.1.11, and 13.1.12.

13.1.22 Calculate and tabulate the normalized linear oscillator wave functions

ψn(x)= 2−n/2π−1/4(n!)−1/2Hn(x)exp

(−x2

2

)for x = 0.0(0.1)5.0

andn= 0(1)5. If a plotting routine is available, plot your results.

13.1.23 Evaluate∫∞−∞ e−2x2

HN1(x) · · ·HN4(x) dx in closed form.

Hint.∫∞−∞e−2x2

HN1(x)HN2(x)HN3(x) dx = 1π

2(N1+N2+N3−1)/2 · Ŵ(s −N1)Ŵ(s −N2)

· Ŵ(s − N3), s = (N1 + N2 + N3 + 1)/2 or∫∞−∞ e−2x2

HN1(x)HN2(x) dx =(−1)(N1+N2−1)/22(N1+N2−1)/2 · Ŵ((N1 + N2 + 1)/2) may be helpful. Prove these for-mulas (see Gradshteyn and Ryzhik, no. 7.375 on p. 844, in the Additional Readings).

13.2 Laguerre Functions 837

13.2 LAGUERRE FUNCTIONS

Differential Equation — Laguerre Polynomials

If we start with the appropriate generating function, it is possible to develop the Laguerrepolynomials in analogy with the Hermite polynomials. Alternatively, a series solution maybe developed by the methods of Section 9.5. Instead, to illustrate a different technique, letus start with Laguerre’s ODE and obtain a solution in the form of a contour integral, as wedid with the integral representation for the modified Bessel functionKν(x) (Section 11.6).From this integral representation a generating function will be derived.

Laguerre’s ODE (which derives from the radial ODE of Schrödinger’s PDE for the hy-drogen atom) is

xy′′(x)+ (1− x)y′(x)+ ny(x)= 0. (13.52)

We shall attempt to representy, or ratheryn, sincey will depend on the parametern,a nonnegative integer, by the contour integral

yn(x)=1

2πi

∮e−xz/(1−z)

(1− z)zn+1dz (13.53a)

and demonstrate that it satisfies Laguerre’s ODE. The contour includes the origin but doesnot enclose the pointz= 1. By differentiating the exponential in Eq. (13.53a) we obtain

y′n(x) = −1

2πi

∮e−xz/(1−z)

(1− z)2zndz, (13.53b)

y′′n(x) =1

2πi

∮e−xz/(1−z)

(1− z)3zn−1dz. (13.53c)

Substituting into the left-hand side of Eq. (13.52), we obtain

1

2πi

∮ [x

(1− z)3zn−1− 1− x

(1− z)2zn+ n

(1− z)zn+1

]e−xz/(1−z) dz,

which is equal to

− 1

2πi

∮d

dz

[e−xz/(1−z)

(1− z)zn

]dz. (13.54)

If we integrate our perfect differential around a closed contour (Fig. 13.3), the integral willvanish, thus verifying thatyn(x) (Eq. (13.53a)) is a solution of Laguerre’s equation.

It has become customary to defineLn(x), the Laguerre polynomial (Fig. 13.4), by5

Ln(x)=1

2πi

∮e−xz/(1−z)

(1− z)zn+1dz. (13.55)

5Other definitions ofLn(x) are in use. The definitions here of the Laguerre polynomialLn(x) and the associated LaguerrepolynomialLk

n(x) agree with AMS-55, Chapter 22. (For the full ref. see footnote 4 in Chapter 5 or the General References atbook’s end.)


FIGURE 13.3 Laguerrepolynomial contour.

FIGURE 13.4 Laguerrepolynomials.

This is exactly what we would obtain from the series

g(x, z)= e−xz/(1−z)

1− z=

∞∑

n=0

Ln(x)zn, |z|< 1, (13.56)

if we multipliedg(x, z) by z−n−1 and integrated around the origin. As in the developmentof the calculus of residues (Section 7.1), only thez−1 term in the series survives. On thisbasis we identifyg(x, z) as the generating function for the Laguerre polynomials.


With the transformation

xz

1− z= s − x, or z= s − x

s, (13.57)

Ln(x) =ex

2πi

∮sne−s

(s − x)n+1ds, (13.58)

the new contour enclosing the points = x in the s-plane. By Cauchy’s integral formula(for derivatives),

Ln(x)=ex

n!dn

dxn

(xne−x

)(integraln), (13.59)

giving Rodrigues’ formula for Laguerre polynomials. From these representations ofLn(x)

we find the series form (for integraln),

Ln(x) =(−1)n

n!

[xn − n2

1! xn−1+ n2(n− 1)2

2! xn−2− · · · + (−1)nn!]

=n∑

m=0

(−1)mn!xm(n−m)!m!m! =

n∑

s=0

(−1)n−sn!xn−s(n− s)!(n− s)!s! (13.60)

and the specific polynomials listed in Table 13.2 (Exercise 13.2.1). Clearly, the defini-tion of Laguerre polynomials in Eqs. (13.55), (13.56), (13.59), and (13.60) are equivalent.Practical applications will decide which approach is used as one’s starting point. Equa-tion (13.59) is most convenient for generating Table 13.2, Eq. (13.56) for deriving recur-sion relations from which the ODE (13.52) is recovered.

By differentiating the generating function in Eq. (13.56) with respect tox and z, weobtain recurrence relations for the Laguerre polynomials as follows. Using the productrule for differentiation we verify the identities

(1− z)2∂g

∂z= (1− x − z)g(x, z), (z− 1)

∂g

∂x= zg(x, z). (13.61)

Table 13.2 Laguerre Polynomials

L0(x)= 1L1(x)=−x + 12!L2(x)= x2− 4x + 23!L3(x)=−x3+ 9x2− 18x + 64!L4(x)= x4− 16x3+ 72x2− 96x + 245!L5(x)=−x5+ 25x4− 200x3+ 600x2− 600x + 1206!L6(x)= x6− 36x5+ 450x4− 2400x3+ 5400x2− 4320x + 720


Writing the left-hand and right-hand sides of the first identity in terms of Laguerre polyno-mials using Eq. (13.56) we obtain

∑

n

[(n+ 1)Ln+1(x)− 2nLn(x)+ (n− 1)Ln−1(x)

]zn

=∑

n

[(1− x)Ln(x)−Ln−1(x)

]zn.

Equating coefficients ofzn yields

(n+ 1)Ln+1(x)= (2n+ 1− x)Ln(x)− nLn−1(x). (13.62)

To get the second recursion relation we use both identities of Eqs. (13.61) to verify thethird identity,

x∂g

∂x= z

∂g

∂z− z

∂(zg)

∂z, (13.63)

which, when written similarly in terms of Laguerre polynomials, is seen to be equivalentto

xL′n(x)= nLn(x)− nLn−1(x). (13.64)

Equation (13.61), modified to read

Ln+1(x)= 2Ln(x)−Ln−1(x)−1

n+ 1

[(1+ x)Ln(x)−Ln−1(x)

], (13.65)

for reasons of economy and numerical stability, is used for computation of numerical val-ues ofLn(x). The computer starts with known numerical values ofL0(x) andL1(x), Ta-ble 13.2, and works up step by step. This is the same technique discussed for computingLegendre polynomials, Section 12.2.

Also, from Eq. (13.56) we find

g(0, z)= 1

1− z=

∞∑

n=0

zn =∞∑

n=0

Ln(0)zn,

which yields the special values of Laguerre polynomials

Ln(0)= 1. (13.66)

As is seen from the form of the generating function, from the form of Laguerre’s ODE, orfrom Table 13.2, the Laguerre polynomials have neither odd nor even symmetry under theparity transformationx→−x.

The Laguerre ODE is not self-adjoint, and the Laguerre polynomialsLn(x) do not bythemselves form an orthogonal set. However, following the method of Section 10.1, if wemultiply Eq. (13.52) bye−x (Exercise 10.1.1) we obtain

∫ ∞

0e−xLm(x)Ln(x) dx = δmn. (13.67)


This orthogonality is a consequence of the Sturm–Liouville theory, Section 10.1. The nor-malization follows from the generating function. It is sometimes convenient to define or-thogonalized Laguerre functions (with unit weighting function) by

ϕn(x)= e−x/2Ln(x). (13.68)

Our new orthonormal function,ϕn(x), satisfies the ODE

xϕ′′n(x)+ ϕ′n(x)+(n+ 1

2− x

4

)ϕn(x)= 0, (13.69)

which is seen to have the (self-adjoint) Sturm–Liouville form. Note that the interval(0≤ x <∞) was used because Sturm–Liouville boundary conditions are satisfied at itsendpoints.

Associated Laguerre Polynomials

In many applications, particularly in quantum mechanics, we need the associated Laguerrepolynomials defined by6

Lkn(x)= (−1)k

dk

dxkLn+k(x). (13.70)

From the series form ofLn(x) we verify that the lowest associated Laguerre polynomialsare given by

Lk0(x) = 1,

Lk1(x) = −x + k+ 1,

Lk2(x) =

x2

2− (k + 2)x + (k + 2)(k + 1)

2. (13.71)

In general,

Lkn(x)=

n∑

m=0

(−1)m(n+ k)!

(n−m)!(k +m)!m!xm, k >−1. (13.72)

A generating function may be developed by differentiating the Laguerre generating func-tion k times to yield

(−1)kdk

dxk

e−xz/(1−z)

1− z= (−1)k

∞∑

n=0

dk

dxkLn+k(x)zn+k =

∞∑

n=0

Lkn(x)z

n+k

=(

z

1− z

)kexz/(1−z)

1− z.

6Some authors useLkn+k(x)= (dk/dxk)[Ln+k(x)]. Hence ourLk

n(x)= (−1)kLkn+k(x).


From the last two members of this equation, canceling the common factorzk , we obtain

e−xz/(1−z)

(1− z)k+1=

∞∑

n=0

Lkn(x)z

n, |z|< 1. (13.73)

From this, forx = 0, the binomial expansion

1

(1− z)k+1=

∞∑

n=0

(−k− 1

n

)(−z)n =

∞∑

n=0

Lkn(0)z

n

yields

Lkn(0)=

(n+ k)!n!k! . (13.74)

Recurrence relations can be derived from the generating function or by differentiating theLaguerre polynomial recurrence relations. Among the numerous possibilities are

(n+ 1)Lkn+1(x) = (2n+ k + 1− x)Lk

n(x)− (n+ k)Lkn−1(x), (13.75)

xdLk

n(x)

dx= nLk

n(x)− (n+ k)Lkn−1(x). (13.76)

Thus, we obtain from differentiating Laguerre’s ODE once

xdL′′ndx

+L′′n −L′n + (1− x)dL′ndx

+ ndLn

dx= 0,

and eventually from differentiating Laguerre’s ODEk times

xdk

dxkL′′n + k

dk−1

dxk−1L′′n − k

dk−1

dxk−1L′n + (1− x)

dk

dxkL′n + n

dk

dxkLn = 0.

Adjusting the indexn→ n+ k, we have the associated Laguerre ODE

xd2Lk

n(x)

dx2+ (k + 1− x)

dLkn(x)

dx+ nLk

n(x)= 0. (13.77)

When associated Laguerre polynomials appear in a physical problem it is usually becausethat physical problem involves Eq. (13.77). The most important application is the boundstates of the hydrogen atom, which are derived in upcoming Example 13.2.1.

A Rodrigues representation of the associated Laguerre polynomial

Lkn(x)=

exx−k

n!dn

dxn

(e−xxn+k

)(13.78)

may be obtained from substituting Eq. (13.59) into Eq. (13.70). Note that all these formulasfor associated Legendre polynomialsLk

n(x) reduce to the corresponding expressions forLn(x) whenk = 0.


The associated Laguerre equation (13.77) is not self-adjoint, but it can be put in self-adjoint form by multiplying bye−xxk , which becomes the weighting function (Sec-tion 10.1). We obtain

∫ ∞

0e−xxkLk

n(x)Lkm(x) dx =

(n+ k)!n! δmn. (13.79)

Equation (13.79) shows the same orthogonality interval(0,∞) as that for the Laguerrepolynomials, but with a new weighting function we have a new set of orthogonal polyno-mials, the associated Laguerre polynomials.

By lettingψkn(x)= e−x/2xk/2Lk

n(x),ψkn(x) satisfies the self-adjoint ODE

xd2ψk

n(x)

dx2+ dψk

n(x)

dx+(−x

4+ 2n+ k+ 1

2− k2

4x

)ψkn(x)= 0. (13.80)

Theψkn(x) are sometimes calledLaguerre functions. Equation (13.67) is the special case

k = 0 of Eq. (13.79).A further useful form is given by defining7

�kn(x)= e−x/2x(k+1)/2Lk

n(x). (13.81)

Substitution into the associated Laguerre equation yields

d2�kn(x)

dx2+(−1

4+ 2n+ k + 1

2x− k2− 1

4x2

)�k

n(x)= 0. (13.82)

The corresponding normalization integral∫∞

0 |�kn(x)|2dx is

∫ ∞

0e−xxk+1[Lk

n(x)]2dx = (n+ k)!

n! (2n+ k + 1). (13.83)

Notice that the�kn(x) donot form an orthogonal set (except withx−1 as a weighting func-

tion) because of thex−1 in the term(2n+ k + 1)/2x. (The Laguerre functionsLµν (x) in

which the indicesν andµ arenot integers may be defined using the confluent hypergeo-metric functions of Section 13.5.)

Example 13.2.1 THE HYDROGEN ATOM

The most important application of the Laguerre polynomials is in the solution of theSchrödinger equation for the hydrogen atom. This equation is

− h2

2m∇2ψ − Ze2

4πǫ0rψ =Eψ, (13.84)

in which Z = 1 for hydrogen, 2 for ionized helium, and so on. Separating variables, wefind that the angular dependence ofψ is the spherical harmonicsYM

L (θ,ϕ). The radialpart,R(r), satisfies the equation

− h2

2m

1

r2

d

dr

(r2dR

dr

)− Ze2

4πǫ0rR + h2

2m

L(L+ 1)

r2R =ER. (13.85)

7This corresponds to modifying the functionψ in Eq. (13.80) to eliminate the first derivative (compare Exercise 9.6.11).


For bound states,R→ 0 asr→∞, andR is finite at the origin,r = 0. We do not considercontinuum states with positive energy. Only when the latter are included do hydrogen wavefunctions form a complete set.

By use of the abbreviations (resulting from rescalingr to the dimensionless radial vari-ableρ)

ρ = αr with α2=−8mE

h2, E < 0, λ= mZe2

2πǫ0αh2, (13.86)

Eq. (13.85) becomes

1

ρ2

d

dρ

(ρ2dχ(ρ)

dρ

)+(λ

ρ− 1

4− L(L+ 1)

ρ2

)χ(ρ)= 0, (13.87)

whereχ(ρ)= R(ρ/α). A comparison with Eq. (13.82) for�kn(x) shows that Eq. (13.87)

is satisfied by

ρχ(ρ)= e−ρ/2ρL+1L2L+1λ−L−1(ρ), (13.88)

in which k is replaced by 2L+ 1 andn by λ−L− 1, upon using

1

ρ2

d

dρρ2dχ

dρ= 1

ρ

d2

dρ2(ρχ).

We must restrict the parameterλ by requiring it to be an integern, n= 1,2,3, . . . .8 Thisis necessary because the Laguerre function of nonintegraln would diverge9 asρneρ , whichis unacceptable for our physical problem, in which

limr→∞

R(r)= 0.

This restriction onλ, imposed by our boundary condition, has the effect of quantizing theenergy,

En =−Z2m

2n2h2

(e2

4πǫ0

)2

. (13.89)

The negative sign reflects the fact that we are dealing here with bound states (E < 0), cor-responding to an electron that is unable to escape to infinity, where the Coulomb potentialgoes to zero. Using this result forEn, we have

α = me2

2πǫ0h2· Zn= 2Z

na0, ρ = 2Z

na0r, (13.90)

with

a0=4πǫ0h

2

me2, the Bohr radius.

8This is the conventional notation forλ. It is not the samen as the indexn in �kn(x).

9This may be shown, as in Exercise 9.5.5.


Thus, the final normalized hydrogen wave function is written as

ψnLM(r, θ,ϕ)=[(

2Z

na0

)3(n−L− 1)!2n(n+L)!

]1/2

e−αr/2(αr)LL2L+1n−L−1(αr)Y

ML (θ,ϕ).

(13.91)

Regular solutions exist forn ≥ L + 1, so the lowest state withL = 1 (called a 2P state)occurs only withn= 2. �

Exercises

13.2.1 Show with the aid of the Leibniz formula that the series expansion ofLn(x)

(Eq. (13.60)) follows from the Rodrigues representation (Eq. (13.59)).

13.2.2 (a) Using the explicit series form (Eq. (13.60)) show that

L′n(0) = −n,L′′n(0) = 1

2n(n− 1).

(b) Repeat without using the explicit series form ofLn(x).

13.2.3 From the generating function derive the Rodrigues representation

Lkn(x)=

exx−k

n!dn

dxn

(e−xxn+k

).

13.2.4 Derive the normalization relation (Eq. (13.79)) for the associated Laguerre polynomials.

13.2.5 Expandxr in a series of associated Laguerre polynomialsLkn(x), k fixed andn ranging

from 0 tor (or to∞ if r is not an integer).Hint. The Rodrigues form ofLk

n(x) will be useful.

ANS. xr = (r + k)!r!r∑

n=0

(−1)nLkn(x)

(n+ k)!(r − n)! , 0≤ x <∞.

13.2.6 Expande−ax in a series of associated Laguerre polynomialsLkn(x), k fixed andn rang-

ing from 0 to∞.

(a) Evaluate directly the coefficients in your assumed expansion.(b) Develop the desired expansion from the generating function.

ANS. e−ax = 1

(1+ a)1+k

∞∑

n=0

(a

1+ a

)n

Lkn(x), 0≤ x <∞.

13.2.7 Show that ∫ ∞

0e−xxk+1Lk

n(x)Lkn(x) dx =

(n+ k)!n! (2n+ k + 1).

Hint. Note that

xLkn = (2n+ k + 1)Lk

n − (n+ k)Lkn−1− (n+ 1)Lk

n+1.


13.2.8 Assume that a particular problem in quantum mechanics has led to the ODE

d2y

dx2−[k2− 1

4x2− 2n+ k + 1

2x+ 1

4

]y = 0

for nonnegative integersn, k. Write y(x) as

y(x)=A(x)B(x)C(x),

with the requirement that

(a) A(x) be anegativeexponential giving the required asymptotic behavior ofy(x)

and(b) B(x) be apositivepower ofx giving the behavior ofy(x) for 0≤ x≪ 1.

DetermineA(x) andB(x). Find the relation betweenC(x) and the associated Laguerrepolynomial.

ANS.A(x)= e−x/2, B(x)= x(k+1)/2, C(x)= Lkn(x).

13.2.9 From Eq. (13.91) the normalized radial part of the hydrogenic wave function is

RnL(r)=[α3 (n−L− 1)!

2n(n+L)!

]1/2

e−αr(αr)LL2L+1n−L−1(αr),

in whichα = 2Z/na0= 2Zme2/4πε0h2. Evaluate

(a) 〈r〉 =∫ ∞

0rRnL(αr)RnL(αr)r

2 dr ,

(b)⟨r−1⟩=

∫ ∞

0r−1RnL(αr)RnL(αr)r

2 dr .

The quantity〈r〉 is the average displacement of the electron from the nucleus, whereas〈r−1〉 is the average of the reciprocal displacement.

ANS. 〈r〉 = a0

2

[3n2−L(L+ 1)

],

⟨r−1⟩= 1

n2a0.

13.2.10 Derive the recurrence relation for the hydrogen wave function expectation values:

s + 2

n2

⟨rs+1⟩− (2s + 3)a0

⟨rs⟩+ s + 1

4

[(2L+ 1)2− (s + 1)2

]a2

0

⟨rs−1⟩= 0,

with s ≥−2L− 1, 〈rs〉 ≡ $rs .Hint. Transform Eq. (13.87) into a form analogous to Eq. (13.80). Multiply byρs+2u′−cρs+1u. Hereu= ρ�. Adjustc to cancel terms that do not yield expectation values.

13.2.11 The hydrogen wave functions, Eq. (13.91), are mutually orthogonal, as they should be,since they are eigenfunctions of the self-adjoint Schrödinger equation

∫ψ∗n1L1M1

ψn2L2M2r2dr d�= δn1n2δL1L2δM1M2.


Yet the radial integral has the (misleading) form∫ ∞

0e−αr/2(αr)LL2L+1

n1−L−1(αr)e−αr/2(αr)LL2L+1

n2−L−1(αr)r2 dr,

which appearsto match Eq. (13.83) and not the associated Laguerre orthogonality re-lation, Eq. (13.79). How do you resolve this paradox?

ANS. The parameterα is dependent onn. The first threeα, previ-ously shown, are 2Z/n1a0. The last three are 2Z/n2a0. Forn1= n2 Eq. (13.83) applies. Forn1 = n2 neither Eq. (13.79)nor Eq. (13.83) is applicable.

13.2.12 A quantum mechanical analysis of the Stark effect (parabolic coordinate) leads to theODE

d

dξ

(ξdu

dξ

)+(

1

2Eξ +L− m2

4ξ− 1

4Fξ2

)u= 0.

HereF is a measure of the perturbation energy introduced by an external electric field.Find the unperturbed wave functions(F = 0) in terms of associated Laguerre polyno-mials.

ANS.u(ξ)= e−εξ/2ξm/2Lmp (εξ), with ε =

√−2E > 0,

p = L/ε− (m+ 1)/2, a nonnegative integer.

13.2.13 The wave equation for the three-dimensional harmonic oscillator is

− h2

2M∇2ψ + 1

2Mω2r2ψ =Eψ.

Hereω is the angular frequency of the corresponding classical oscillator. Show that theradial part ofψ (in spherical polar coordinates ) may be written in terms of associatedLaguerre functions of argument(βr2), whereβ =Mω/h.Hint. As in Exercise 13.2.8, split off radial factors ofr l ande−βr

2/2. The associatedLaguerre function will have the formLl+1/2

1/2(n−l−1)(βr2).

13.2.14 Write a computer program that will generate the coefficientsas in the polynomial formof the Laguerre polynomialLn(x)=

∑ns=0asx

s .

13.2.15 Write a computer program that will transform a finite power series∑N

n=0anxn into a

Laguerre series∑N

n=0bnLn(x). Use the recurrence relation, Eq. (13.62).

13.2.16 TabulateL10(x) for x = 0.0(0.1)30.0. This will include the 10 roots ofL10. Beyondx = 30.0,L10(x) is monotonic increasing. If graphic software is available, plot yourresults.

Check value.Eighth root= 16.279.

13.2.17 Determine the 10 roots ofL10(x) using root-finding software. You may use your knowl-edge of the approximate location of the roots or develop a search routine to look for theroots. The 10 roots ofL10(x) are the evaluation points for the 10-point Gauss–Laguerrequadrature. Check your values by comparing with AMS-55, Table 25.9. (For the refer-ence see footnote 4 in Chapter 5 or the General References at book’s end.)


13.2.18 Calculate the coefficients of a Laguerre series expansion(Ln(x), k = 0) of the ex-ponentiale−x . Evaluate the coefficients by the Gauss–Laguerre quadrature (compareEq. (10.64)). Check your results against the values given in Exercise 13.2.6.Note. Direct application of the Gauss–Laguerre quadrature withf (x)Ln(x)e

−x givespoor accuracy because of the extrae−x . Try a change of variable,y = 2x, so that thefunction appearing in the integrand will be simplyLn(y/2).

13.2.19 (a) Write a subroutine to calculate the Laguerre matrix elements

Mmnp =∫ ∞

0Lm(x)Ln(x)x

pe−x dx.

Include a check of the condition|m− n| ≤ p ≤m+ n. (If p is outside this range,Mmnp = 0. Why?)Note. A 10-point Gauss–Laguerre quadrature will give accurate results form+ n+ p ≤ 19.

(b) Call your subroutine to calculate a variety of Laguerre matrix elements. CheckMmn1 against Exercise 13.2.7.

13.2.20 Write a subroutine to calculate the numerical value ofLkn(x) for specified values ofn, k,

andx. Require thatn andk be nonnegative integers andx ≥ 0.Hint. Starting with known values ofLk

0 andLk1(x), we may use the recurrence relation,

Eq. (13.75), to generateLkn(x), n= 2,3,4, . . . .

13.2.21 Show that∫∞−∞ xne−x

2Hn(xy) dx =

√πn!Pn(y), wherePn is a Legendre polynomial.

13.2.22 Write a program to calculate the normalized hydrogen radial wave functionψnL(r).This isψnLM of Eq. (13.91), omitting the spherical harmonicYM

L (θ,ϕ). TakeZ = 1anda0 = 1 (which means thatr is being expressed in units of Bohr radii). Acceptn

andL as input data. TabulateψnL(r) for r = 0.0(0.2)R with R taken large enough toexhibit the significant features ofψ . This means roughlyR = 5 for n= 1, R = 10 forn= 2, andR = 30 forn= 3.

13.3 CHEBYSHEV POLYNOMIALS

In this section two types of Chebyshev polynomials are developed as special cases of ul-traspherical polynomials. Their properties follow from the ultraspherical polynomial gen-erating function. The primary importance of the Chebyshev polynomials is in numericalanalysis.

Generating Functions

In Section 12.1 the generating function for the ultraspherical, or Gegenbauer, polynomials

1

(1− 2xt + t2)α=

∞∑

n=0

C(α)n (x)tn, |x|< 1, |t |< 1 (13.92)

was mentioned, withα = 12 giving rise to the Legendre polynomials. In this section we first

takeα = 1 and thenα = 0 to generate two sets of polynomials known as the Chebyshevpolynomials.

13.3 Chebyshev Polynomials 849

Type II

With α = 1 andC(1)n (x)=Un(x), Eq. (13.92) gives

1

1− 2xt + t2=

∞∑

n=0

Un(x)tn, |x|< 1, |t |< 1. (13.93)

These functionsUn(x) generated by(1− 2xt + t2)−1 are labeled Chebyshev polynomialstype II. Although these polynomials have few applications in mathematical physics, oneunusual application is in the development of four-dimensional spherical harmonics used inangular momentum theory.

Type I

With α = 0 there is a difficulty. Indeed, our generating function reduces to the constant 1.We may avoid this problem by first differentiating Eq. (13.92) with respect tot . This yields

−α(−2x + 2t)

(1− 2xt + t2)α+1=

∞∑

n=1

nC(α)n (x)tn−1, (13.94)

or

x − t

(1− 2xt + t2)α+1=

∞∑

n=1

n

2

[C

(α)n (x)

α

]tn−1. (13.95)

We defineC(0)n (x) by

C(0)n (x)= lim

α→0

C(α)n (x)

α. (13.96)

The purpose of differentiating with respect tot was to getα in the denominator andto create an indeterminate form. Now multiplying Eq. (13.95) by 2t and adding 1=(1− 2xt + t2)/(1− 2xt + t2), we obtain

1− t2

1− 2xt + t2= 1+ 2

∞∑

n=1

n

2C(0)n (x)tn. (13.97)

We defineTn(x) by

Tn(x)={

1, n= 0n

2C(0)n (x), n > 0.

(13.98)

Notice the special treatment forn= 0. This is similar to the treatment of then= 0 term inthe Fourier series. Also, note thatC

(0)n is the limit indicated in Eq. (13.96) and not a literal

substitution ofα = 0 into the generating function series. With these new labels,

1− t2

1− 2xt + t2= T0(x)+ 2

∞∑

n=1

Tn(x)tn, |x| ≤ 1, |t |< 1. (13.99)


We callTn(x) the type I Chebyshev polynomials. Note that the notation and spelling of thename for these functions differs from reference to reference. Here we follow the usage ofAMS-55 (for the full reference see footnote 4 in Chapter 5).

Differentiating the generating function (Eqs. (13.99)) with respect tot and multiplyingby the denominator, 1− 2xt + t2, we obtain

−t − (t − x)

[T0(x)+ 2

∞∑

n=1

Tn(x)tn

]=(1− 2xt + t2

) ∞∑

n=1

nTn(x)tn−1

=∞∑

n=1

[nTnt

n−1− 2xnTntn + nTnt

n+1],

from which the recurrence relation

Tn+1(x)− 2xTn(x)+ Tn−1(x)= 0 (13.100)

follows by shifting the summation index so as to get the same power,tn, in each term andthen comparing coefficients oftn. Similarly treating Eq. (13.93) we find

− 2(t − x)

1− 2xt + t2=(1− 2xt + t2

) ∞∑

n=1

nUn(x)tn−1

from which the recursion relation

Un+1(x)− 2xUn(x)+Un−1(x)= 0 (13.101)

follows upon comparing coefficients of like powers oft (see Table 13.3).Then, using the generating functions for the first few values ofn and these recurrence

relations for the higher-order polynomials, we get Tables 13.4 and 13.5 (see also Figs. 13.5and 13.6).

As with the Hermite polynomials, Section 13.1, the recurrence relations, Eqs. (13.100)and (13.101), together with the known values ofT0(x), T1(x),U0(x), andU1(x), provide aconvenient — that is, for a computer — means of getting the numerical value of anyTn(x0)

or Un(x0), with x0 a given number.

Table 13.3 Recursion relationa Pn+1(x)=(Anx +Bn)Pn(x)−CnPn−1(x)

Pn(x) An Bn Cn

Legendre Pn(x)2n+1n+1 0 1

n+1Chebyshev I Tn(x) 2 0 1Shifted Chebyshev I T ∗n (x) 4 −2 1Chebyshev II Un(x) 2 0 1Shifted Chebyshev II U∗n (x) 4 −2 1

Associated Laguerre L(k)n (x) − 1

n+12n+k+1n+1

n+kn+1

Hermite Hn(x) 2 0 2n

aPn denotes any of the orthogonal polynomials.


Table 13.4 Chebyshevpolynomials, type I

T0= 1T1= x

T2= 2x2− 1T3= 4x3− 3xT4= 8x4− 8x2+ 1T5= 16x5− 20x3+ 5xT6= 32x6− 48x4+ 18x2− 1

Table 13.5 Chebyshevpolynomials, type II

U0= 1U1= 2xU2= 4x2− 1U3= 8x3− 4xU4= 16x4− 12x2+ 1U5= 32x5− 32x3+ 6xU6= 64x6− 80x4+ 24x2− 1

FIGURE 13.5 Chebyshev polynomialsTn(x).

FIGURE 13.6 Chebyshev polynomialsUn(x).


Again, from the generating functions, we can obtain the special values of various poly-nomials:

Tn(1) = 1, Tn(−1)= (−1)n,(13.102)

T2n(0) = (−1)n, T2n+1(0)= 0;

Un(1) = n+ 1, Un(−1)= (−1)n(n+ 1),(13.103)

U2n(0) = (−1)n, U2n+1(0)= 0.

For example, comparing the power series

1− t2

(1− t)2= 1+ t

1− t=

∞∑

n=0

(tn + tn+1)

with Eq. (13.99) forx = 1 givesTn(1), and forx =−1 a similar expansion of(1− t)/(1+t) givesTn(−1), while replacingt→−t2 in the first power series yieldsTn(0). The powerseries for(1± t)−2 and(1+ t2)−1 generate the correspondingUn(±1),Un(0).

The parity relations forTn andUn follow from their generating functions, with the sub-stitutionst→−t, x→−x, which leave them invariant; these are

Tn(x)= (−1)nTn(−x), Un(x)= (−1)nUn(−x). (13.104)

Rodrigues’ representations ofTn(x) andUn(x) are

Tn(x)=(−1)nπ1/2(1− x2)1/2

2n(n− 12)!

dn

dxn

[(1− x2)n−1/2] (13.105)

and

Un(x)=(−1)n(n+ 1)π1/2

2n+1(n+ 12)!(1− x2)1/2

dn

dxn

[(1− x2)n+1/2]

. (13.106)

Recurrence Relations — Derivatives

Differentiation of the generating functions forTn(x) andUn(x) with respect to the variablex leads to a variety of recurrence relations involving derivatives. For example, from Eq.(13.99) we thus obtain

(1− 2xt + t2

)2∞∑

n=1

T ′n(x)tn = 2t

[T0(x)+ 2

∞∑

n=1

Tn(x)tn

],

from which we extract the recursion

2Tn−1(x)= T ′n(x)− 2xT ′n−1(x)+ T ′n−2(x), (13.107)

which is the derivative of Eq. (13.100) forn→ n− 1. Among the more useful recursionswe thus find are

(1− x2)T ′n(x)=−nxTn(x)+ nTn−1(x) (13.108)


and(1− x2)U ′n(x)=−nxUn(x)+ (n+ 1)Un−1(x). (13.109)

Manipulating a variety of these recursions as in Section 12.2 for Legendre polynomials onecan eliminate the indexn− 1 also in favor ofT ′′n and establish thatTn(x), the Chebyshevpolynomial type I, satisfies the ODE

(1− x2)T ′′n (x)− xT ′n(x)+ n2Tn(x)= 0. (13.110)

The Chebyshev polynomial of type II,Un(x), satisfies(1− x2)U ′′n (x)− 3xU ′n(x)+ n(n+ 2)Un(x)= 0. (13.111)

Chebyshev polynomials may be defined starting from these ODEs, but our emphasis hasbeen on generating functions.

The ultraspherical equation

(1− x2) d2

dx2C(α)n (x)− (2α + 1)x

d

dxC(α)n (x)+ n(n+ 2α)C(α)

n (x)= 0 (13.112)

is a generalization of these differential equations, reducing to Eq. (13.110) forα = 0 andEq. (13.111) forα = 1 (and to Legendre’s equation forα = 1

2).

Trigonometric Form

At this point in the development of the properties of the Chebyshev solutions it is beneficialto change variables, replacingx by cosθ . With x = cosθ andd/dx = (−1/sinθ)(d/dθ),we verify that

(1− x2)d2Tn

dx2= d2Tn

dθ2− cotθ

dTn

dθ, xT ′n =−cotθ

dTn

dθ.

Adding these terms, Eq. (13.110) becomes

d2Tn

dθ2+ n2Tn = 0, (13.113)

the simple harmonic oscillator equation with solutions cosnθ and sinnθ . The special val-ues (boundary conditions atx = 0,1) identify

Tn = cosnθ = cosn(arccosx). (13.114a)

A second linearly independent solution of Eqs. (13.110) and (13.113) is labeled

Vn = sinnθ = sinn(arccosx). (13.114b)

The corresponding solutions of the type II Chebyshev equation, Eq. (13.111), become

Un =sin(n+ 1)θ

sinθ, (13.115a)

Wn =cos(n+ 1)θ

sinθ. (13.115b)


The two sets of solutions, type I and type II, are related by

Vn(x) =(1− x2)1/2

Un−1(x), (13.116a)

Wn(x) =(1− x2)−1/2

Tn+1(x). (13.116b)

As already seen from generating functions,Tn(x) andUn(x) are polynomials. Clearly,Vn(x) andWn(x) arenot polynomials. From

Tn(x)+ iVn(x) = cosnθ + i sinnθ

= (cosθ + i sinθ)n =[x + i

(1− x2)1/2]n

, |x| ≤ 1 (13.117)

we obtain expansions

Tn(x)= xn −(n

2

)xn−2(1− x2)+

(n

4

)xn−4(1− x2)2− · · · (13.118a)

and

Vn(x)=√

1− x2

[(n

1

)xn−1−

(n

3

)xn−3(1− x2)+ · · ·

]. (13.118b)

From the generating functions, or from the ODEs, power-series representations are

Tn(x)=n

2

[n/2]∑

m=0

(−1)m(n−m− 1)!m!(n− 2m)! (2x)

n−2m, (13.119a)

for n≥ 1, with [n/2] the largest integer belown/2 and

Un(x)=[n/2]∑

m=0

(−1)m(n−m)!

m!(n− 2m)! (2x)n−2m. (13.119b)

Orthogonality

If Eq. (13.110) is put into self-adjoint form (Section 10.1), we obtainw(x)= (1− x2)−1/2

as a weighting factor. For Eq. (13.111) the corresponding weighting factor is(1− x2)+1/2.The resulting orthogonality integrals,

∫ 1

−1Tm(x)Tn(x)

(1− x2)−1/2

dx =

0, m = n,π

2, m= n = 0,

π, m= n= 0,

(13.120)

∫ 1

−1Vm(x)Vn(x)

(1− x2)−1/2

dx =

0, m = n,π

2, m= n = 0,

0, m= n= 0,

(13.121)

∫ 1

−1Um(x)Un(x)

(1− x2)1/2

dx = π

2δm,n, (13.122)


and∫ 1

−1Wm(x)Wn(x)

(1− x2)1/2

dx = π

2δm,n, (13.123)

are a direct consequence of the Sturm–Liouville theory, Chapter 10. The normalizationvalues may best be obtained by usingx = cosθ and converting these four integrals intoFourier normalization integrals (for the half-period interval[0,π]).

Exercises

13.3.1 Another Chebyshev generating function is

1− xt

1− 2xt + t2=

∞∑

n=0

Xn(x)tn, |t |< 1.

How isXn(x) related toTn(x) andUn(x)?

13.3.2 Given(1− x2)U ′′n (x)− 3xU ′n(x)+ n(n+ 2)Un(x)= 0,

show thatVn(x) (Eq. (13.116a)) satisfies(1− x2)V ′′n (x)− xV ′n(x)+ n2Vn(x)= 0,

which is Chebyshev’s equation.

13.3.3 Show that the Wronskian ofTn(x) andVn(x) is given by

Tn(x)V′n(x)− T ′n(x)Vn(x)=−

n

(1− x2)1/2.

This verifies thatTn andVn(n = 0) are independent solutions of Eq. (13.110). Con-versely, forn= 0, we do not have linear independence. What happens atn= 0? Whereis the “second” solution?

13.3.4 Show thatWn(x)= (1− x2)−1/2Tn+1(x) is a solution of(1− x2)W ′′

n (x)− 3xW ′n(x)+ n(n+ 2)Wn(x)= 0.

13.3.5 Evaluate the Wronskian ofUn(x) andWn(x)= (1− x2)−1/2Tn+1(x).

13.3.6 Vn(x) = (1− x2)1/2Un−1(x) is not defined forn = 0. Show that a second and inde-pendent solution of the Chebyshev differential equation forTn(x) (n = 0) is V0(x) =arccosx (or arcsinx).

13.3.7 Show that Vn(x) satisfies the same three-term recurrence relation asTn(x)

(Eq. (13.100)).

13.3.8 Verify the series solutions forTn(x) andUn(x) (Eqs. (13.109a) and (13.119b)).


13.3.9 Transform the series form ofTn(x), Eq. (13.119a), into anascendingpower series.

ANS. T2n(x)= (−1)nnn∑

m=0

(−1)m(n+m− 1)!(n−m)!(2m)! (2x)

2m, n≥ 1,

T2n+1(x)=2n+ 1

2

n∑

m=0

(−1)m+n(n+m)!(n−m)!(2m+ 1)! (2x)

2m+1.

13.3.10 Rewrite the series form ofUn(x), Eq. (13.119b), as an ascending power series.

ANS.U2n(x)= (−1)nn∑

m=0

(−1)m(n+m)!

(n−m)!(2m)! (2x)2m,

U2n+1(x)= (−1)nn∑

m=0

(−1)m(n+m+ 1)!

(n−m)!(2m+ 1)! (2x)2m+1.

13.3.11 Derive the Rodrigues representation ofTn(x),

Tn(x)=(−1)nπ1/2(1− x2)1/2

2n(n− 12)!

dn

dxn

[(1− x2)n−1/2]

.

Hint. One possibility is to use the hypergeometric function relation

2F1(a, b; c; z)= (1− z)−a 2F1

(a, c− b; c; −z

1− z

),

with z= (1−x)/2. An alternate approach is to develop a first-order differential equationfor y = (1− x2)n−1/2. Repeated differentiation of this equation leads to the Chebyshevequation.

13.3.12 (a) From the differential equation forTn (in self-adjoint form) show that∫ 1

−1

dTm(x)

dx

dTn(x)

dx

(1− x2)1/2

dx = 0, m = n.

(b) Confirm the preceding result by showing that

dTn(x)

dx= nUn−1(x).

13.3.13 The expansion of a power ofx in a Chebyshev series leads to the integral

Imn =∫ 1

−1xmTn(x)

dx√1− x2

.

(a) Show that this integral vanishes form< n.(b) Show that this integral vanishes form+ n odd.

13.3.14 Evaluate the integral

Imn =∫ 1

−1xmTn(x)

dx√1− x2

for m≥ n andm+ n even by each of two methods:


(a) Operate withx as the variable replacingTn by its Rodrigues representation.(b) Usingx = cosθ transform the integral to a form withθ as the variable.

ANS. Imn = πm!

(m− n)!(m− n− 1)!!(m+ n)!! , m≥ n, m+ n even.

13.3.15 Establish the following bounds,−1≤ x ≤ 1:

(a) |Un(x)| ≤ n+ 1, (b)

∣∣∣∣d

dxTn(x)

∣∣∣∣≤ n2.

13.3.16 (a) Establish the following bound,−1≤ x ≤ 1: |Vn(x)| ≤ 1.(b) Show thatWn(x) is unbounded in−1≤ x ≤ 1.

13.3.17 Verify the orthogonality-normalization integrals for

(a)Tm(x), Tn(x), (b) Vm(x), Vn(x),(c) Um(x), Un(x), (d) Wm(x), Wn(x).

Hint. All these can be converted to Fourier orthogonality-normalization integrals.

13.3.18 Show whether

(a) Tm(x) andVn(x) are or are not orthogonal over the interval[−1,1] with respectto the weighting factor(1− x2)−1/2.

(b) Um(x) andWn(x) are or are not orthogonal over the interval[−1,1] with respectto the weighting factor(1− x2)1/2.

13.3.19 Derive

(a) Tn+1(x)+ Tn−1(x)= 2xTn(x),(b) Tm+n(x)+ Tm−n(x)= 2Tm(x)Tn(x),

from the “corresponding” cosine identities.

13.3.20 A number of equations relate the two types of Chebyshev polynomials. As examplesshow that

Tn(x)=Un(x)− xUn−1(x)

and(1− x2)Un(x)= xTn+1(x)− Tn+2(x).

13.3.21 Show that

dVn(x)

dx=−n Tn(x)√

1− x2

(a) using the trigonometric forms ofVn andTn,(b) using the Rodrigues representation.


13.3.22 Starting withx = cosθ andTn(cosθ)= cosnθ , expand

xk =(eiθ + e−iθ

2

)k

and show that

xk = 1

2k−1

[Tk(x)+

(k

1

)Tk−2(x)+

(k

2

)Tk−4+ · · ·

],

the series in brackets terminating with(km

)T1(x) for k = 2m+1 or 1

2

(km

)T0 for k = 2m.

13.3.23 (a) Calculate and tabulate the Chebyshev functionsV1(x),V2(x), andV3(x) for x =−1.0(0.1)1.0.

(b) A second solution of the Chebyshev differential equation, Eq. (13.100), forn = 0 is y(x) = sin−1x. Tabulate and plot this function over the same range:−1.0(0.1)1.0.

13.3.24 Write a computer program that will generate the coefficientsas in the polynomial formof the Chebyshev polynomialTn(x)=

∑ns=0asx

s .

13.3.25 TabulateT10(x) for 0.00(0.01)1.00. This will include the five positive roots ofT10. Ifgraphics software is available, plot your results.

13.3.26 Determine the five positive roots ofT10(x) by calling a root-finding subroutine. Useyour knowledge of the approximate location of these roots from Exercise 13.3.25 orwrite a search routine to look for the roots. These five positive roots (and their negatives)are the evaluation points of the 10-point Gauss–Chebyshev quadrature method.

Check values.xk = cos[(2k − 1)π/20

], k = 1,2,3,4,5.

13.3.27 Develop the following Chebyshev expansions (for[−1,1]):

(a)(1− x2)1/2= 2

π

[1− 2

∞∑

s=1

(4s2− 1

)−1T2s(x)

].

(b)+1, 0< x ≤ 1−1, −1≤ x < 0

}= 4

π

∞∑

s=0

(−1)s(2s + 1)−1T2s+1(x).

13.3.28 (a) For the interval[−1,1] show that

|x| = 1

2+

∞∑

s=1

(−1)s+1 (2s − 3)!!(2s + 2)!! (4s + 1)P2s(x)

= 2

π+ 4

π

∞∑

s=1

(−1)s+1 1

4s2− 1T2s(x).

(b) Show that the ratio of the coefficient ofT2s(x) to that ofP2s(x) approaches(πs)−1

ass→∞. This illustrates the relatively rapid convergence of the Chebyshev se-ries.

13.4 Hypergeometric Functions 859

Hint. With the Legendre recurrence relations, rewritexPn(x) as a linear combinationof derivatives. The trigonometric substitutionx = cosθ,Tn(x)= cosnθ is most helpfulfor the Chebyshev part.

13.3.29 Show that

π2

8= 1+ 2

∞∑

s=1

(4s2− 1

)−2.

Hint. Apply Parseval’s identity (or the completeness relation) to the results of Exer-cise 13.3.28.

13.3.30 Show that

(a) cos−1x = π

2− 4

π

∞∑

n=0

1

(2n+ 1)2T2n+1(x).

(b) sin−1x = 4

π

∞∑

n=0

1

(2n+ 1)2T2n+1(x).

13.4 HYPERGEOMETRIC FUNCTIONS

In Chapter 9 the hypergeometric equation10

x(1− x)y′′(x)+[c− (a + b+ 1)x

]y′(x)− ab y(x)= 0 (13.124)

was introduced as a canonical form of a linear second-order ODE with regular singularitiesatx = 0,1, and∞. One solution is

y(x) = 2F1(a, b; c;x)

= 1+ a · bc

x

1! +a(a + 1)b(b+ 1)

c(c+ 1)

x2

2! + · · · , c = 0,−1,−2,−3, . . . ,

(13.125)

which is known as thehypergeometric function or hypergeometric series. The range ofconvergence forc > a + b is |x|< 1 andx = 1, and isx =−1 for c > a + b− 1. In termsof the often-used Pochhammer symbol,

(a)n = a(a + 1)(a + 2) · · · (a + n− 1)= (a + n− 1)!(a − 1)! ,

(a)0 = 1, (13.126)

the hypergeometric function becomes

2F1(a, b; c;x)=∞∑

n=0

(a)n(b)n

(c)n

xn

n! . (13.127)

10This is sometimes called Gauss’ ODE. The solutions then become Gauss functions.


In this form the subscripts 2 and 1 become clear. The leading subscript 2 indicates thattwo Pochhammer symbols appear in the numerator and the final subscript 1 indicates onePochhammer symbol in the denominator.11 (The confluent hypergeometric function1F1

with one Pochhammer symbol in the numerator and one in the denominator appears inSection 13.5.)

From the form of Eq. (13.125) we see that the parameterc may not be zero or a negativeinteger. On the other hand, ifa or b equals 0 or a negative integer, the series terminates andthe hypergeometric function becomes a polynomial. Many more or less elementary func-tions can be represented by the hypergeometric function.12 Comparing the power series weverify that

ln(1+ x)= x 2F1(1,1;2;−x). (13.128)

For the complete elliptic integralsK andE,

K(k2) =

∫ π/2

0

(1− k2 sin2 θ

)−1/2dθ = π

22F1

(1

2,

1

2;1; k2

), (13.129)

E(k2) =

∫ π/2

0

(1− k2 sinθ

)1/2dθ = π

22F1

(1

2,−1

2;1; k2

). (13.130)

The explicit series forms and other properties of the elliptic integrals are developed inSection 5.8.

The hypergeometric equation as a second-order linear ODE has a second independentsolution. The usual form is

y(x)= x1−c2F1(a + 1− c, b+ 1− c;2− c;x), c = 2,3,4, . . . . (13.131)

If c is an integer either the two solutions coincide or (barring a rescue by integrala orintegral b) one of the solutions will blow up (see Exercise 13.4.1). In such a case thesecond solution is expected to include a logarithmic term.

Alternate forms of the hypergeometric ODE include

(1− z2) d2

dz2y

(1− z

2

)−[(a + b+ 1)z− (a + b+ 1− 2c)

] ddz

y

(1− z

2

)

− ab y

(1− z

2

)= 0, (13.132)

(1− z2) d2

dz2y(z2)−

[(2a + 2b+ 1)z+ 1− 2c

z

]d

dzy(z2)− 4ab y

(z2)= 0. (13.133)

11The Pochhammer symbol is often useful in other expressions involving factorials, for instance,

(1− z)−a =∞∑

n=0

(a)nzn/n!, |z|< 1.

12With three parameters,a, b, andc, we can represent almost anything.

13.4 Hypergeometric Functions 861

Contiguous Function Relations

The parametersa, b, andc enter in the same way as the parametern of Bessel, Legendre,and other special functions. As we found with these functions, we expect recurrence rela-tions involving unit changes in the parametersa, b, andc. The usual nomenclature for thehypergeometric functions, in which one parameter changes by+ or −1, is a “contiguousfunction.” Generalizing this term to include simultaneous unit changes in more than oneparameter, we find 26 functions contiguous to2F1(a, b; c;x). Taking them two at a time,we can develop the formidable total of 325 equations among the contiguous functions. Onetypical example is

(a − b){c(a + b− 1)+ 1− a2− b2+

[(a − b)2− 1

](1− x)

}2F1(a, b; c;x)

= (c− a)(a − b+ 1)b 2F1(a − 1, b+ 1; c;x)+ (c− b)(a − b− 1)a 2F1(a + 1, b− 1; c;x). (13.134)

Another contiguous function relation appears in Exercise 13.4.10.

Hypergeometric Representations

Since the ultraspherical equation (13.112) in Section 13.3 is a special case of Eq. (13.124),we see that ultraspherical functions (and Legendre and Chebyshev functions) may be ex-pressed as hypergeometric functions. For the ultraspherical function we obtain

Cβn (x)=

(n+ 2β)!2βn!β! 2F1

(−n,n+ 2β + 1;1+ β; 1− x

2

)(13.135)

upon comparing its ODE with Eq. (13.124) and the power-series solutions. For Legendreand associated Legendre functions we find similarly

Pn(x) = 2F1

(−n,n+ 1;1; 1− x

2

), (13.136)

Pmn (x) = (n+m)!

(n−m)!(1− x2)m/2

2mm! 2F1

(m− n,m+ n+ 1;m+ 1; 1− x

2

). (13.137)

Alternate forms are

P2n(x) = (−1)n(2n)!

22nn!n! 2F1

(−n,n+ 1

2; 1

2;x2

)

= (−1)n(2n− 1)!!(2n)!! 2F1

(−n,n+ 1

2; 1

2;x2

), (13.138)

P2n+1(x) = (−1)n(2n+ 1)!22nn!n! 2F1

(−n,n+ 3

2; 3

2;x2

)x

= (−1)n(2n+ 1)!!(2n)!! 2F1

(−n,n+ 3

2; 3

2;x2

)x. (13.139)


In terms of hypergeometric functions, the Chebyshev functions become

Tn(x) = 2F1

(−n,n; 1

2; 1− x

2

), (13.140)

Un(x) = (n+ 1) 2F1

(−n,n+ 2; 3

2; 1− x

2

), (13.141)

Vn(x) = n√

1− x22F1

(−n+ 1, n+ 1; 3

2; 1− x

2

). (13.142)

The leading factors are determined by direct comparison of complete power series, com-parison of coefficients of particular powers of the variable, or evaluation atx = 0 or 1, andso on.

The hypergeometric series may be used to define functions with nonintegral indices. Thephysical applications are minimal.

Exercises

13.4.1 (a) Forc, an integer, anda andb nonintegral, show that

2F1(a, b; c;x) and x1−c2F1(a + 1− c, b+ 1− c;2− c;x)

yield only one solution to the hypergeometric equation.(b) What happens ifa is an integer, say,a =−1, andc=−2?

13.4.2 Find the Legendre, Chebyshev I, and Chebyshev II recurrence relations correspondingto the contiguous hypergeometric function Eq. (13.134).

13.4.3 Transform the following polynomials into hypergeometric functions of argumentx2. (a)T2n(x); (b) x−1T2n+1(x); (c) U2n(x); (d) x−1U2n+1(x).

ANS. (a)T2n(x)= (−1)n 2F1(−n,n; 12;x2).

(b) x−1T2n+1(x)= (−1)n(2n+ 1) 2F1(−n,n+ 1; 32;x2).

(c) U2n(x)= (−1)n 2F1(−n,n+ 1; 12;x2).

(d) x−1U2n+1(x)= (−1)n(2n+ 2) 2F1(−n,n+ 2; 32;x2).

13.4.4 Derive or verify the leading factor in the hypergeometric representations of the Cheby-shev functions.

13.4.5 Verify that the Legendre function of the second kind,Qν(z), is given by

Qν(z) =π1/2ν!

(ν + 12)!(2z)ν+1 2F1

(ν

2+ 1

2,ν

2+ 1; ν

2+ 3

2; z−2

),

|z| > 1, |argz|< π, ν = −1,−2,−3, . . . .

13.4.6 Analogous to the incomplete gamma function, we may define an incomplete beta func-tion by

Bx(a, b)=∫ x

0ta−1(1− t)b−1dt.

13.5 Confluent Hypergeometric Functions 863

Show that

Bx(a, b)= a−1xa 2F1(a,1− b;a + 1;x).

13.4.7 Verify the integral representation

2F1(a, b; c; z)=Ŵ(c)

Ŵ(b)Ŵ(c− b)

∫ 1

0tb−1(1− t)c−b−1(1− tz)−a dt.

What restrictions must be placed on the parametersb andc and on the variablez?Note. The restriction on|z| can be dropped — analytic continuation. For nonintegrala

the real axis in thez-plane from 1 to∞ is a cut line.Hint. The integral is suspiciously like a beta function and can be expanded into a seriesof beta functions.

ANS.ℜ(c) >ℜ(b) > 0, and|z|< 1.

13.4.8 Prove that

2F1(a, b; c;1)=Ŵ(c)Ŵ(c− a − b)

Ŵ(c− a)Ŵ(c− b), c = 0,−1,−2, . . . c > a + b.

Hint. Here is a chance to use the integral representation, Exercise 13.4.7.

13.4.9 Prove that

2F1(a, b; c;x)= (1− x)−a 2F1

(a, c− b; c; −x

1− x

).

Hint. Try an integral representation.Note. This relation is useful in developing a Rodrigues representation ofTn(x) (compareExercise 13.3.11).

13.4.10 Verify that

2F1(−n,b; c;1)=(c− b)n

(c)n.

Hint. Here is a chance to use the contiguous function relation[2a − c + (b − a)x] ·2F1(a, b; c;x)= a(1− x) 2F1(a + 1, b; c;x)− (c− a) 2F1(a − 1, b; c;x) and mathe-matical induction. Alternatively, use the integral representation and the beta function.

13.5 CONFLUENT HYPERGEOMETRIC FUNCTIONS

The confluent hypergeometric equation13

xy′′(x)+ (c− x)y′(x)− ay(x)= 0 (13.143)

13This is often calledKummer’s equation. The solutions, then, areKummer functions.


has a regular singularity atx = 0 and an irregular one atx =∞. It is obtained from the hy-pergeometric equation of Section 13.4 by merging (by hand:x(1−x)→ x in Eq. (13.124))two of the latter’s three singularities. One solution of the confluent hypergeometric equa-tion is

y(x) = 1F1(a; c;x)=M(a, c, x)

= 1+ a

c

x

1! +a(a + 1)

c(c+ 1)

x2

2! + · · · , c = 0,−1,−2, . . . . (13.144)

This solution is convergent for all finitex (or complexz). In terms of the Pochhammersymbols, we have

M(a, c, x)=∞∑

n=0

(a)n

(c)n

xn

n! . (13.145)

Clearly,M(a, c, x) becomes a polynomial if the parametera is 0 or a negative integer.Numerous more or less elementary functions may be represented by the confluent hyper-geometric function. Examples are the error function and the incomplete gamma function(from Eq. (8.69)):

erf(x) = 2

π1/2

∫ x

0e−t

2dt = 2

π1/2xM

(1

2,

3

2,−x2

), (13.146)

γ (a, x) =∫ x

0e−t ta−1dt = a−1xaM(a,a + 1,−x), ℜ(a) > 0. (13.147)

Clearly, this coincides with the first solution forc = a. The error function and the incom-plete gamma function are discussed further in Section 8.5.

A second solution of Eq. (13.143) is given by

y(x)= x1−cM(a + 1− c,2− c, x), c = 2,3,4, . . . . (13.148)

The standard form of the second solution of Eq. (13.143) is a linear combination ofEqs. (13.144) and (13.148):

U(a, c, x)= π

sinπc

[M(a, c, x)

(a − c)!(c− 1)! −x1−cM(a + 1− c,2− c, x)

(a − 1)!(1− c)!

]. (13.149)

Note the resemblance to our definition of the Neumann function, Eq. (11.60). As with ourNeumann function, Eq. (11.60), this definition ofU(a, c, x) becomes indeterminate in thiscase forc an integer.

An alternate form of the confluent hypergeometric equation that will be useful later isobtained by changing the independent variable fromx to x2:

d2

dx2y(x2)+

[2c− 1

x− 2x

]d

dxy(x2)− 4ay

(x2)= 0. (13.150)

As with the hypergeometric functions, contiguous functions exist in which the para-metersa and c are changed by±1. Including the cases of simultaneous changes in the


two parameters,14 we have eight possibilities. Taking the original function and pairs of thecontiguous functions, we can develop a total of 28 equations.15

Integral Representations

It is frequently convenient to have the confluent hypergeometric functions in integral form.We find (Exercise 13.5.10)

M(a, c, x) = Ŵ(c)

Ŵ(a)Ŵ(c− a)

∫ 1

0ext ta−1(1− t)c−a−1dt, ℜ(c) >ℜ(a) > 0,

(13.151)

U(a, c, x) = 1

Ŵ(a)

∫ ∞

0e−xt ta−1(1+ t)c−a−1dt, ℜ(x) > 0, ℜ(a) > 0.

(13.152)

Three important techniques for deriving or verifying integral representations are as fol-lows:

1. Transformation of generating function expansions and Rodrigues representations: TheBessel and Legendre functions provide examples of this approach.

2. Direct integration to yield a series: This direct technique is useful for a Bessel functionrepresentation (Exercise 11.1.18) and a hypergeometric integral (Exercise 13.4.7).

3. (a) Verification that the integral representation satisfies the ODE. (b) Exclusion ofthe other solution. (c) Verification of normalization. This is the method used in Sec-tion 11.5 to establish an integral representation of the modified Bessel functionKν(z).It will work here to establish Eqs. (13.151) and (13.152).

Bessel and Modified Bessel Functions

Kummer’s first formula,

M(a, c, x)= exM(c− a, c,−x), (13.153)

is useful in representing the Bessel and modified Bessel functions. The formula may be ver-ified by series expansion or by use of an integral representation (compare Exercise 13.5.10).

As expected from the form of the confluent hypergeometric equation and the characterof its singularities, the confluent hypergeometric functions are useful in representing anumber of the special functions of mathematical physics. For the Bessel functions,

Jν(x)=e−ix

ν!

(x

2

)ν

M

(ν + 1

2,2ν + 1,2ix

), (13.154)

whereas for the modified Bessel functions of the first kind,

Iν(x)=e−x

ν!

(x

2

)ν

M

(ν + 1

2,2ν + 1,2x

). (13.155)

14Slater refers to these asassociated functions.15The recurrence relations for Bessel, Hermite, and Laguerre functions are special cases of these equations.


Hermite Functions

The Hermite functions are given by

H2n(x) = (−1)n(2n)!n! M

(−n, 1

2, x2

), (13.156)

H2n+1(x) = (−1)n2(2n+ 1)!

n! xM

(−n, 3

2, x2

), (13.157)

using Eq. (13.150).Comparing the Laguerre ODE with the confluent hypergeometric equation (13.143), we

have

Ln(x)=M(−n,1, x). (13.158)

The constant is fixed as unity by noting Eq. (13.66) forx = 0. For the associated Laguerrefunctions,

Lmn (x)= (−1)m

dm

dxmLn+m(x)=

(n+m)!n!m! M(−n,m+ 1, x). (13.159)

Alternate verification is obtained by comparing Eq. (13.159) with the power-series so-lution (Eq. (13.72) of Section 13.2). Note that in the hypergeometric form, as distinct froma Rodrigues representation, the indicesn andm need not be integers, and, if they are notintegers,Lm

n (x) will not be a polynomial.

Miscellaneous Cases

There are certain advantages in expressing our special functions in terms of hypergeomet-ric and confluent hypergeometric functions. If the general behavior of the latter functions isknown, the behavior of the special functions we have investigated follows as a series of spe-cial cases. This may be useful in determining asymptotic behavior or evaluating normaliza-tion integrals. The asymptotic behavior ofM(a, c, x) andU(a, c, x) may be convenientlyobtained from integral representations of these functions, Eqs. (13.151) and (13.152). Thefurther advantage is that the relations between the special functions are clarified. For in-stance, an examination of Eqs. (13.156), (13.157), and (13.159) suggests that the Laguerreand Hermite functions are related.

The confluent hypergeometric equation (13.143) is clearly not self-adjoint. For this andother reasons it is convenient to define

Mkµ(x)= e−x/2xµ+1/2M

(µ− k+ 1

2,2µ+ 1, x

). (13.160)

This new function,Mkµ(x), is a Whittaker function that satisfies the self-adjoint equation

M ′′kµ(x)+

(−1

4+ k

x+

14 −µ2

x2

)Mkµ(x)= 0. (13.161)

The corresponding second solution is

Wkµ(x)= e−x/2xµ+1/2U

(µ− k + 1

2,2µ+ 1, x

). (13.162)


Exercises

13.5.1 Verify the confluent hypergeometric representation of the error function

erf(x)= 2x

π1/2M

(1

2,

3

2,−x2

).

13.5.2 Show that the Fresnel integralsC(x) andS(x) of Exercise 5.10.2 may be expressed interms of the confluent hypergeometric function as

C(x)+ iS(x)= xM

(1

2,

3

2,iπx2

2

).

13.5.3 By direct differentiation and substitution verify that

y = ax−a∫ x

0e−t ta−1dt = ax−aγ (a, x)

satisfies

xy′′ + (a + 1+ x)y′ + ay = 0.

13.5.4 Show that the modified Bessel function of the second kind,Kν(x), is given by

Kν(x)= π1/2e−x(2x)νU(ν + 1

2,2ν + 1,2x

).

13.5.5 Show that the cosine and sine integrals of Section 8.5 may be expressed in terms ofconfluent hypergeometric functions as

Ci(x)+ i si(x)=−eixU(1,1,−ix).This relation is useful in numerical computation of Ci(x) and si(x) for large values ofx.

13.5.6 Verify the confluent hypergeometric form of the Hermite polynomialH2n+1(x)

(Eq. (13.157)) by showing that

(a) H2n+1(x)/x satisfies the confluent hypergeometric equation witha = −n, c = 32

and argumentx2,

(b) limx→0

H2n+1(x)

x= (−1)n

2(2n+ 1)!n! .

13.5.7 Show that the contiguous confluent hypergeometric function equation

(c− a)M(a − 1, c, x)+ (2a − c+ x)M(a, c, x)− aM(a + 1, c, x)= 0

leads to the associated Laguerre function recurrence relation (Eq. (13.75)).

13.5.8 Verify the Kummer transformations:

(a) M(a, c, x)= exM(c− a, c,−x)(b) U(a, c, x)= x1−cU(a − c+ 1,2− c, x).


13.5.9 Prove that

(a)dn

dxnM(a, c, x)= (a)n

(b)nM(a + n,b+ n,x),

(b)dn

dxnU(a, c, x)= (−1)n(a)nU(a + n, c+ n,x).

13.5.10 Verify the following integral representations:

(a) M(a, c, x)= Ŵ(c)

Ŵ(a)Ŵ(c− a)

∫ 1

0ext ta−1(1− t)c−a−1dt, ℜ(c) >ℜ(a) > 0.

(b) U(a, c, x)= 1

Ŵ(a)

∫ ∞

0e−xt ta−1(1+ t)c−a−1dt, ℜ(x) > 0, ℜ(a) > 0.

Under what conditions can you acceptℜ(x)= 0 in part (b)?

13.5.11 From the integral representation ofM(a, c, x), Exercise 13.5.10(a), show that

M(a, c, x)= exM(c− a, c,−x).Hint. Replace the variable of integrationt by 1− s to release a factorex from theintegral.

13.5.12 From the integral representation ofU(a, c, x), Exercise 13.5.10(b), show that the expo-nential integral is given by

E1(x)= e−xU(1,1, x).

Hint. Replace the variable of integrationt in E1(x) by x(1+ s).

13.5.13 From the integral representations ofM(a, c, x) andU(a, c, x) in Exercise 13.5.10 de-velop asymptotic expansions of(a)M(a, c, x), (b) U(a, c, x).Hint. You can use the technique that was employed withKν(z), Section 11.6.

ANS. (a)Ŵ(c)

Ŵ(a)

ex

xc−a

{1+ (1− a)(c− a)

1!x +

(1− a)(2− a)(c− a)(c− a + 1)

2!x2+ · · ·

}

(b)1

xa

{1+ a(1+ a − c)

1!(−x) + a(a + 1)(1+ a − c)(2+ a − c)

2!(−x)2 + · · ·}

.

13.5.14 Show that the Wronskian of the two confluent hypergeometric functionsM(a, c, x) andU(a, c, x) is given by

MU ′ −M ′U =− (c− 1)!(a − 1)!

ex

xc.

What happens ifa is 0 or a negative integer?

13.6 Mathieu Functions 869

13.5.15 The Coulomb wave equation (radial part of the Schrödinger equation with Coulombpotential) is

d2y

dρ2+[1− 2η

ρ− L(L+ 1)

ρ2

]y = 0.

Show that a regular solutiony = FL(η,ρ) is given by

FL(η,ρ)= CL(η)ρL+1e−iρM(L+ 1− iη,2L+ 2,2iρ).

13.5.16 (a) Show that the radial part of the hydrogen wave function, Eq. (13.81), may be writ-ten as

e−αr/2(αr)LL2L+1n−L−1(αr)

= (n+L)!(n−L− 1)!(2L+ 1)!e

−αr/2(αr)LM(L+ 1− n,2L+ 2, αr).

(b) It was assumed previously that the total (kinetic+ potential) energyE of the elec-tron was negative. Rewrite the (unnormalized) radial wave function for the freeelectron,E > 0.

ANS. eiαr/2(αr)LM(L + 1− in,2L + 2,−iαr), outgoing wave.This representation provides a powerful alternative tech-nique for the calculation of photoionization and recombina-tion coefficients.

13.5.17 Evaluate

(a)∫ ∞

0

[Mkµ(x)

]2dx, (b)

∫ ∞

0

[Mkµ(x)

]2dxx

,

(c)∫ ∞

0

[Mkµ(x)

]2 dx

x1−a ,

where 2µ= 0,1,2, . . . , k −µ− 12 = 0,1,2, . . . , a >−2µ− 1.

ANS. (a)(2µ)!2k. (b) (2µ)!. (c) (2µ)!(2k)a .

13.6 MATHIEU FUNCTIONS

When PDEs such as Laplace’s, Poisson’s, and the wave equation are solved with cylin-drical or spherical boundary conditions by separating variables in polar coordinates, wefind radial solutions, which are the Bessel functions of Chapter 11, and angular solutions,which are sinmϕ,cosmϕ in cylindrical cases and spherical harmonics in spherical cases.Examples are electromagnetic waves in resonant cavities, vibrating circular drumheads,and coaxial wave guides.

When in such cylindrical problems the circular boundary condition becomes ellipticalwe are led to the angular and radial Mathieu functions, which, therefore, might be calledelliptic cylinder functions. In fact, in 1868 Mathieu developed the leading terms of seriessolutions of the vibrating elliptical drumhead, and Whittaker and others in the early 1900sderived higher-order terms as well.


Here our goal is to give an introduction to the rich and complex properties of Mathieufunctions.

Separation of Variables in Elliptical Coordinates

Elliptical cylinder coordinatesξ, η, z, which are appropriate for elliptical boundary condi-tions, are expressed in rectangular coordinates as

x = c coshξ cosη, y = c sinhξ sinη, z= z, (13.163)

0≤ ξ <∞, 0≤ η ≤ 2π,

where the parameter 2c > 0 is the distance between the foci of the confocal ellipses de-scribed by these coordinates (Fig. 13.7). We want to show that in the limitc→ 0 the fociof the ellipses coalesce to the center of circles. We work at constantz-coordinate mostly,z= 0, say. Indeed for fixed radial variableξ =const. we can eliminate the angular variableη to obtain from Eq. (13.163)

x2

c2 cosh2 ξ+ y2

c2 sinh2 ξ= 1, (13.164)

describing confocal ellipses centered at the origin of thex, y-plane with major and minorhalf-axes

a = c coshξ, b= c sinhξ, (13.165)

respectively. Since

b

a= tanhξ =

√1− 1

cosh2 ξ≡√

1− e2, (13.166)

the eccentricitye = 1/coshξ of the ellipse with 0≤ e ≤ 1, and the distance between thefoci 2ae = 2c, providing a geometrical interpretation of the radial coordinateξ and the

FIGURE 13.7 Elliptical coordinatesξ, η.


parameterc. As ξ →∞, e→ 0 and the ellipses become circles, which is indicated inFig. 13.7. Asξ → 0, the ellipse becomes more elongated until, atξ = 0, it has shrunk tothe line segment between the foci.

Whenη=const. we eliminateξ to find confocal hyperbolas

x2

c2 cos2η− y2

c2 sin2η= 1, (13.167)

which are also plotted in Fig. 13.7. Differentiating the ellipse, we obtain

x dx

cosh2 ξ+ y dy

sinh2 ξ= 0, (13.168)

which means that the tangent vector(dx, dy) of the ellipse is perpendicular to the vector( x

cosh2 ξ,

y

sinh2 ξ). For the hyperbola the orthogonality condition is

x dx

cos2η− y dy

sin2η= 0, (13.169)

so the scalar product of the ellipse and hyperbola tangent vectors at each of their intersec-tion points(x, y) of Eq. (13.163) obey

x2

cosh2 ξ cos2η− y2

sinh2 ξ sin2η= c2− c2= 0. (13.170)

This means that these confocal ellipses and hyperbolas form an orthogonal coordinatesystem, in the sense of Section 2.1. To extract the scale factorshξ , hη from the differentialsof the elliptical coordinates

dx = c sinhξ cosη dξ − c coshξ sinη dη,(13.171)

dy = c coshξ sinη dξ + c sinhξ cosη dη,

we sum their squares, finding

dx2+ dy2 = c2(sinh2 ξ cos2η+ cosh2 ξ sin2η)(dξ2+ dη2)

= c2(cosh2 ξ − cos2η)(dξ2+ dη2)≡ h2

ξ dξ2+ h2

η dη2 (13.172)

and yielding

hξ = hη = c(cosh2 ξ − cos2η

)1/2. (13.173)

Note that there is no cross term involvingdξ dη, showing again that we are dealing withorthogonal coordinates.

Now we are ready to derive Mathieu’s differential equations.


Example 13.6.1 ELLIPTICAL DRUM

We consider vibrations of an elliptical drumhead with vertical displacementz= z(x, y, t)

governed by the wave equation

∂2z

∂x2+ ∂2z

∂y2= 1

v2

∂2z

∂t2, (13.174)

where the velocity squaredv2= T/ρ with tensionT and mass densityρ is a constant. Wefirst separate the harmonic time dependence, writing

z(x, y, t)= u(x, y)w(t), (13.175)

wherew(t)= cos(ωt + δ), with ω the frequency andδ a constant phase. Substituting thisfunctionz into Eq. (13.174) yields

1

u

(∂2u

∂x2+ ∂2u

∂y2

)= 1

v2w

∂2w

∂t2=−ω2

v2=−k2= const., (13.176)

that is, the two-dimensional Helmholtz equation for the displacementu. We now use Eq.(2.22) to convert the Laplacian∇2 to the elliptical coordinates, where we drop thez-coordinate. This gives

∂2u

∂x2+ ∂2u

∂y2+ k2u= 1

h2ξ

(∂2u

∂ξ2+ ∂2u

∂η2

)+ k2u= 0, (13.177)

that is, the Helmholtz equation in ellipticalξ, η coordinates,

∂2u

∂ξ2+ ∂2u

∂η2+ c2k2(cosh2 ξ − cos2η

)u= 0. (13.178)

Lastly, we separateξ andη, writing u(ξ, η)=R(ξ)�(η), which yields

1

R

d2R

dξ2+ c2k2 cosh2 ξ = c2k2 cos2η− 1

�

d2�

dη2= λ+ 1

2c2k2, (13.179)

whereλ + c2k2/2 is the separation constant. Writing cosh 2ξ,cos2η instead of cosh2 ξ,cos2η (which motivates the special form of the separation constant in Eq. (13.179)) wefind the linear, second-order ODE

d2R

dξ2− (λ− 2q cosh 2ξ)R(ξ)= 0, q = 1

4c2k2, (13.180)

which is also called theradial Mathieu equation, and

d2�

dη2+ (λ− 2q cos2η)�(η)= 0, (13.181)

theangular, or modified, Mathieu equation. Note that the eigenvalueλ(q) is a functionof the continuous parameterq in the Mathieu ODEs. It is this parameter dependence thatcomplicates the analysis of Mathieu functions and makes them among the most difficultspecial functions used in physics. �


Clearly, all finite points are regular points of both ODEs, while infinity is an essentialsingularity for both ODEs, which are of the Sturm–Liouville type (Chapter 10) with coef-ficient functionsp ≡ 1 and

q(ξ)=−λ+ 2q cosh 2ξ, q(η)= λ− 2q cos2η. (13.182)

(These functionsq must not be confused with the parameterq.) As a consequence, theirsolutions form orthogonal sets of functions. The substitutionη→ iξ transforms the angularto the radial Mathieu ODE, so their solutions are closely related.

Using the Lindemann–Stieltjes substitutionz = cos2η, dz/dη = −sin 2η, the angularMathieu ODE is transformed into an ODE with coefficients that are algebraic in the vari-

ablez (using ddη= dz

dηddz=−sin 2η d

dzand d2

dη2 =−2 cos2η ddz+ sin2 2η d2

dz2 ):

4z(1− z)d2�

dz2+ 2(1− 2z)

d�

dz+[λ+ 2q(1− 2z)

]�= 0. (13.183)

This ODE has regular singularities atz = 0 andz = 1, whereas the point at infinity isan essential singularity (Chapter 9). By comparison, the hypergeometric ODE has threeregular singularities. But not all ODEs with two regular singularities and one essentialsingularity can be transformed into an ODE of the Mathieu type.

Example 13.6.2 THE QUANTUM PENDULUM

A plane pendulum of lengthl and massm with gravitational potentialV (θ)=−mgl cosθis called aquantum pendulum if its wave function� obeys the Schrödinger equation

− h2

2ml2

d2�

dθ2+[V (θ)−E

]� = 0, (13.184)

where the variableθ is the angular displacement from the vertical direction. (For fur-ther details and illustrations we refer to Gutiérrez-Vegaet al. in the Additional Readings.)A boundary condition applies to� so as to be single-valued; that is,�(θ + 2π)=�(θ).Substituting

θ = 2η, λ= 8Eml2

h2, q =−4m2gl3

h2(13.185)

into the Schrödinger equation yields the angular Mathieu ODE for�(2(η + π)) =�(2η). �

For many other applications involving Mathieu functions we refer to Ruby in the Addi-tional Readings.

Our main focus will be on the solutions of the angular Mathieu ODE, which has theimportant property that its coefficient function is periodic with periodπ .


General Properties of Mathieu Functions

In physics applications the angular Mathieu functions are required to be single-valued, thatis, periodic with period 2π . Let us start with some nomenclature. Since Mathieu’s ODEsare invariant under parity (η→−η), Mathieu functions have definite parity. Those of oddparity that have period 2π and, for smallq, start with sin(2n+1)η are called se2n+1(η, q),with n an integer,n = 0,1,2, . . . (se is short for sine-elliptic). Mathieu functions of oddparity and periodπ that start with sin2nη for small q are called se2n(η, q), with n =1,2, . . . . Mathieu functions of even parity, periodπ that start with cos2nη for small qare called ce2n(η, q) (ce is short for cosine-elliptic), while those with period 2π that startwith cos(2n+1)η,n= 0,1, . . . , for smallq are called ce2n+1(η, q). In the limit where theparameterq→ 0 (and the Mathieu ODE becomes the classical harmonic oscillator ODE),Mathieu functions reduce to these trigonometric functions.

The periodicity condition�(η + 2π) = �(η) is sufficient to determine a set of eigen-valuesλ in terms ofq. An elementary analog of this result is the fact that a solution of theclassical harmonic oscillator ODEu′′(η)+λu(η)= 0 has period 2π if, and only if,λ= n2

is the square of an integer. Such problems will be pursued in Section 14.7 as applicationsof Fourier series.

Example 13.6.3 RADIAL MATHIEU FUNCTIONS

Upon replacing the angular elliptic variableη → iξ , the angular Mathieu ODE,Eq. (13.181), becomes the radial ODE, Eq. (13.180). This motivates the definitions ofradial Mathieu functions as

Ce2n+p(ξ, q) = ce2n+p(iξ, q), p = 0,1; n= 0,1, . . . ,

Se2n+p(ξ, q) = −ise2n+p(iξ, q), p = 0,1; n= 1,2, . . . .

Because these functions are differentiable, they correspond to the regular solutions of theradial Mathieu ODE. Of course, they are no longer periodic but are oscillatory (Fig. 13.8).

In physical problems involving elliptical coordinates, the radial Mathieu ODE,Eq. (13.180), plays a role corresponding to Bessel’s ODE in cylindrical geometry. Be-cause there are four families of independent Bessel functions — the regular solutionsJnand irregular Neumann functionsNn, along with the modified Bessel functionsIn andKn — we expect four kinds of radial Mathieu functions. Because of parity, the solutionssplit into even and odd Mathieu functions and so there are eight kinds. Forq > 0,

Je2n(ξ, q) = Ce2n(ξ, q), Je2n+1(ξ, q)=Ce2n+1(ξ, q),

Jo2n(ξ, q) = Se2n(ξ, q), Jo2n+1(ξ, q)= Se2n+1(ξ, q), regular or first kind;Nen(ξ, q),Non(ξ, q), irregular or second kind;

for q < 0, the solutions of the radial Mathieu ODE are denoted by

Ien(ξ, q), Ion(ξ, q), regular or first kind,

Ken(ξ, q),Kon(ξ, q), irregular or second kind


FIGURE 13.8 Radial Mathieu functions:q = 1 (solid line),q = 2 (dashedline), q = 3 (dotted line). (From Gutiérrez-Vegaet al., Am. J. Phys.71:

233 (2003).)

and are known as theevanescent radial Mathieu functions. Mathieu functions corre-sponding to the Hankel functions can be similarly defined. In Fig. 13.8 some of them areplotted.

In applications such as a vibrating drumhead with elliptical boundary conditions (seeExample 13.6.1), the solution can be expanded in even and odd Mathieu functions:

zen ≡ Jen(ξ, q)cen(η, q)cos(ωnt), m≥ 0,

zon ≡ Jon(ξ, q)sen(η, q)cos(ωnt), m≥ 1.

They obey Dirichlet boundary conditions,zen(ξ0, η, t)= 0= zon(ξ0, η, t), which hold pro-vided the radial functions satisfy Jen(ξ0, q) = 0= Jon(ξ0, q) at the elliptical boundary,whereξ = ξ0.

When the focal distancec→ 0, the angular Mathieu functions become the conventionaltrigonometric functions, while the radial Mathieu functions become Bessel functions.

In the case of oscillations of a confocal annular elliptic lake, the modes have to includethe Mathieu functions of the second kind and are thus given by

zen ≡[AJen(ξ, q)+BNen(ξ, q)

]cen(η, q)cos(ωnt), m≥ 0,

zon ≡[AJon(ξ, q)+BNon(ξ, q)

]sen(η, q)cos(ωnt), m≥ 1,

with A,B constants. These standing wave solutions must obey Neumann boundary con-ditions at the inner (ξ = ξ0) and outer (ξ = ξ1) elliptical boundaries; that is, the normal


derivatives (a prime denotesd/dξ ) of zen and zon vanish at each point of the boundaries.For even modes, we have ze′n(ξ0, η, t) = 0= ze′n(ξ1, η, t). The implied radial constraintsare similar to Eqs. (11.81) and (11.82) of Example 11.3.1. Numerical examples and plots,also for traveling waves, are given in Gutiérrez-Vegaet al. in the Additional Readings.�

For zeros of Mathieu functions, their asymptotic expansions, and a more complete listingof formulas we refer to Abramowitz and Stegun (AMS-55) in the Additional Readings,Am.J. Phys.71, Jahnke and Emde and Gradshteyn and Ryzhik in the Additional Readings.

To illustrate and support the nomenclature, we want to show16 that there is an angularMathieu function that is

• even inη and of periodπ if and only if �′1(π/2)= 0;• odd and of periodπ if and only if �2(π/2)= 0;• even and of period 2π if and only if �1(π/2)= 0;• odd and of period 2π if and only if �′2(π/2)= 0,

where�1(η),�2(η) are two linearly independent solutions of the angular Mathieu ODEso that

�1(0)= 1, �′1(0)= 0; �2(0)= 0, �′2(0)= 1. (13.186)

Since the Mathieu ODE is a linear second-order ODE, we know (Chapter 9) that theseinitial conditions are realistic. The first case just given corresponds to ce2n(η, q), with�′1(π/2) = −2nsin2nη|η=π/2 + · · · = 0 for n = 1,2, . . . . The second is the se2n(η, q),with �2(π/2)= sin2nη|π/2+ · · · = 0. The third case is the ce2n+1(η, q), with �1(π/2)=cos(2n+ 1)π/2+ · · · = 0. The fourth case is the se2n+1(η, q).

The key to the proof is Floquet’s approach to linear second-order ODEs with periodiccoefficient functions, such as Mathieu’s angular ODE or the simple pendulum (Exercise13.6.1). If �1(η),�2(η) are two linearly independent solutions of the ODE, any othersolution� can be expressed as

�(η)= c1�1(η)+ c2�2(η), (13.187)

with constantsc1, c2. Now,�k(η+2π) are also solutions because such an ODE is invariantunder the translationη→ η+ 2π , and in particular

�1(η+ 2π) = a1�1(η)+ a2�2(η),

�2(η+ 2π) = b1�1(η)+ b2�2(η), (13.188)

with constantsai, bj . Substituting Eq. (13.188) into Eq. (13.187) we get

�(η+ 2π)= (c1a1+ c2b1)�1(η)+ (c2b2+ c1a2)�2(η), (13.189)

where the constantsci can be chosen as solutions of the eigenvalue equations

a1c1+ b1c2 = λc1,

a2c1+ b2c2 = λc2. (13.190)

16See Hochstadt in the Additional Readings.


ThenFloquet’s theoremstates that�(η+ 2π)= λ�(η), whereλ is a root of∣∣∣∣a1− λ b1a2 b2− λ

∣∣∣∣= 0. (13.191)

A useful corollary is obtained if we defineµ and y by λ = exp(2πµ) and y(η) =exp(−µη)�(η), so

y(η+ 2π)= e−µηe−2πµ�(η+ 2π)= e−µη�(η)= y(η). (13.192)

Thus,�(η)= eµηy(η), with y a periodic function ofη with period 2π .Let us apply Floquet’s argument to the�k(η+π), which are also solutions of Mathieu’s

ODE because the latter is invariant under the translationη→ η+ π . Using the special values in Eq. (13.186) we know that

�1(η+ π) = �1(π)�1(η)+�′1(π)�2(η),

�2(η+ π) = �2(π)�1(η)+�′2(π)�2(η), (13.193)

because these linear combinations of�k(η) are solutions of Mathieu’s ODE with the cor-rect values�i(η+ π),�′i(η+ π) for η= 0. Therefore,

�i(η+ π)= λi�i(η), (13.194)

where theλi are the roots of∣∣∣∣�1(π)− λ �2(π)

�′1(π) �′2(π)− λ

∣∣∣∣= 0. (13.195)

The constant term in the characteristic polynomial is given by the Wronskian

W(�1(η),�2(η)

)= C, (13.196)

a constant because the coefficient ofd�/dη in the angular Mathieu ODE vanishes, imply-ing dW/dη= 0. In fact, using Eq. (13.186),

W(�1(0),�2(0)

)= �1(0)�

′2(0)−�′1(0)�2(0)= 1

=W(�1(π),�2(π)

), (13.197)

so the eigenvalue Eq. (13.195) forλ becomes(�1(π)− λ

)(�′2(π)− λ

)−�2(π)�

′1(π)= 0

= λ2−[�1(π)+�′2(π)

]λ+ 1, (13.198)

with λ1 · λ2= 1 andλ1+ λ2=�1(π)+�′2(π).If |λ1| = |λ2| = 1, thenλ1= exp(iφ) andλ2= exp(−iφ), soλ1+λ2= 2 cosφ. Forφ =

0,π,2π, . . . this case corresponds to|�1(π)+�′2(π)| < 2, where both solutions remainbounded asη→∞ in steps ofπ using Eq. (13.194). These cases do not yield periodicMathieu functions, and this is also the case when|�1(π) + �′2(π)| > 2. If φ = 0, thatis, λ1= 1= λ2 is a double root, then the�i have periodπ and|�1(π)+�′2(π)| = 2. Ifφ = π , that is,λ1=−1= λ2 is again a double root, then|�1(π)+�′2(π)| = −2 and the�i have period 2π with �i(η+ π)=−�i(η).


Because the angular Mathieu ODE is invariant under a parity transformationη→−η,it is convenient to consider solutions

�e(η)=1

2

[�(η)+�(−η)

], �o(η)=

1

2

[�(η)−�(−η)

](13.199)

of definite parity, which obey the same initial conditions as�i . We now relabel�e →�1,�o → �2, taking�1 to be even and�2 to be odd under parity. These solutions ofdefinite parity of Mathieu’s ODE are calledMathieu functions and are labeled accordingto our nomenclature discussed earlier.

If �1(η) has periodπ , then�′1(η + π) = �′1(η) also has periodπ but is odd underparity. Substitutingη=−π/2 we obtain

�′1

(π

2

)=�′1

(−π

2

)=−�′1

(π

2

), so �′1

(π

2

)= 0. (13.200)

Conversely, if�′1(π/2)= 0, then�1(η) has periodπ . To see this, we use

�1(η+ π)= c1�1(η)+ c2�2(η). (13.201)

This expansion is valid because�1(η+ π) is a solution of the angular Mathieu ODE. Wenow determine the coefficientsci , settingη = −π/2, and recall that�1 and�′2 are evenunder parity, whereas�2 and�′1 are odd. This yields

�1

(π

2

)= c1�1

(π

2

)− c2�2

(π

2

),

(13.202)

�′1

(π

2

)= −c1�

′1

(π

2

)+ c2�

′2

(π

2

).

Since�′1(π/2) = 0, �′2(π/2) = 0, or the Wronskian would vanish and�2 ∼ �1 wouldfollow. Hencec2 = 1 follows from the second equation andc1 = 1 from the first. Thus,�1(η+ π)=�1(η). The other bulleted cases listed earlier can be proved similarly.

Because the Mathieu ODEs are of the Sturm–Liouville type, Mathieu functions repre-sent orthogonal systems of functions. So, form,n nonnegative integers, the orthogonalityrelations and normalizations are

∫ π

−πcemcen dη =

∫ π

−πsemsen dη= 0, if m = n;

∫ π

−πcemsen dη = 0; (13.203)

∫ π

−π[ce2n]2dη =

∫ π

−π[se2n]2dη= π, if n≥ 1;

∫ π

0

[ce0(η, q)

]2dη= π.

If a functionf (η) is periodic with periodπ , then it can be expanded in a series of orthog-onal Mathieu functions as

f (η)= 1

2a0 ce0(η, q)+

∞∑

n=1

[an ce2n(η, q)+ bn se2n(η, q)

](13.204)

13.6 Additional Readings 879

with

an =1

π

∫ π

−πf (η)ce2n(η, q) dη, n≥ 0;

(13.205)

bn =1

π

∫ π

−πf (η)se2n(η, q) dη, n≥ 1.

Similar expansions exist for functions of period 2π in terms of ce2n+1 and se2n+1.Series expansions of Mathieu functions will be derived in Section 14.7.

Exercises

13.6.1 For the simple pendulum ODE of Section 5.8, apply Floquet’s method and derive theproperties of its solutions similar to those marked by bullets before Eq. (13.186).

13.6.2 Derive a Mathieu function analog for the Rayleigh expansion of a plane wave forcos(k cosη cosθ) and sin(k cosη cosθ).

Additional Readings

Abramowitz, M., and I. A. Stegun, eds.,Handbook of Mathematical Functions, Applied Mathematics Series-55 (AMS-55). Washington, DC: National Bureau of Standards (1964). Paperback edition, New York: Dover(1974). Chapter 22 is a detailed summary of the properties and representations of orthogonal polynomials.Other chapters summarize properties of Bessel, Legendre, hypergeometric, and confluent hypergeometricfunctions and much more.

Buchholz, H.,The Confluent Hypergeometric Function. New York: Springer-Verlag (1953); translated (1969).Buchholz strongly emphasizes the Whittaker rather than the Kummer forms. Applications to a variety of othertranscendental functions.

Erdelyi, A., W. Magnus, F. Oberhettinger, and F. G. Tricomi,Higher Transcendental Functions, 3 vols. NewYork: McGraw-Hill (1953). Reprinted Krieger (1981). A detailed, almost exhaustive listing of the propertiesof the special functions of mathematical physics.

Fox, L. and I. B. Parker,Chebyshev Polynomials in Numerical Analysis. Oxford: Oxford University Press (1968).A detailed, thorough, but very readable account of Chebyshev polynomials and their applications in numericalanalysis.

Gradshteyn, I. S., and I. M. Ryzhik,Table of Integrals, Series and Products, New York: Academic Press (1980).

Gutiérrez-Vega, J. C., R. M. Rodríguez-Dagnino, M. A. Meneses-Nava and S. Chávez-Cerda,Am. J. Phys.71:233 (2003).

Hochstadt, H.,Special Functions of Mathematical Physics. New York: Holt, Rinehart and Winston (1961),reprinted Dover (1986).

Jahnke, E., and F. Emde,Table of Functions. Leipzig: Teubner (1933); New York: Dover (1943).

Lebedev, N. N.,Special Functions and their Applications(translated by R. A. Silverman). Englewood Cliffs, NJ:Prentice-Hall (1965). Paperback, New York: Dover (1972).

Luke, Y. L.,The Special Functions and Their Approximations. New York: Academic Press (1969). Two volumes:Volume 1 is a thorough theoretical treatment of gamma functions, hypergeometric functions, confluent hy-pergeometric functions, and related functions. Volume 2 develops approximations and other techniques fornumerical work.

Luke, Y. L., Mathematical Functions and Their Approximations. New York: Academic Press (1975). This isan updated supplement toHandbook of Mathematical Functions with Formulas, Graphs and MathematicalTables(AMS-55).


Mathieu, E.,J. de Math. Pures et Appl.13: 137–203 (1868).

McLachlan, N. W.,Theory and Applications of Mathieu Functions. Oxford, UK: Clarendon Press (1947).

Magnus, W., F. Oberhettinger, and R. P. Soni,Formulas and Theorems for the Special Functions of MathematicalPhysics. New York: Springer (1966). An excellent summary of just what the title says, including the topics ofChapters 10 to 13.

Rainville, E. D.,Special Functions. New York: Macmillan (1960), reprinted Chelsea (1971). This book is acoherent, comprehensive account of almost all the special functions of mathematical physics that the reader islikely to encounter.

Rowland, D. R.,Am. J. Phys.72: 758–766 (2004).

Ruby, L.,Am. J. Phys.64: 39–44 (1996).

Sansone, G.,Orthogonal Functions(translated by A. H. Diamond). New York: Interscience (1959). ReprintedDover (1991).

Slater, L. J.,Confluent Hypergeometric Functions. Cambridge, UK: Cambridge University Press (1960). This isa clear and detailed development of the properties of the confluent hypergeometric functions and of relationsof the confluent hypergeometric equation to other ODEs of mathematical physics.

Sneddon, I. N.,Special Functions of Mathematical Physics and Chemistry, 3rd ed. New York: Longman (1980).

Whittaker, E. T., and G. N. Watson,A Course of Modern Analysis. Cambridge, UK: Cambridge University Press,reprinted (1997). The classic text on special functions and real and complex analysis.

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